2019牛客多校第4场

训练记录

D题做了太长时间，一直在找规律，其实lfw一开始就有接近正解的想法，不过还是想要找规律快速过掉，直到找规律找不到以后才开始从二进制每一位的奇偶开始考虑。

题目链接：https://ac.nowcoder.com/acm/contest/884#question

题解

A meeting

题解：https://blog.csdn.net/liufengwei1/article/details/97557656

  1 #include<bits/stdc++.h>
  2 #define maxl 100010
  3 
  4 using namespace std;
  5 const int inf=2e9;
  6 
  7 int n,k,cnt,ans;
  8 int ehead[maxl];
  9 int f[maxl],mxfa[maxl],mxs[maxl],secmxs[maxl];
 10 bool mxflag[maxl],secmxflag[maxl],faflag[maxl];
 11 struct ed
 12 {
 13     int to,nxt,mi;
 14 }e[maxl<<1];
 15 bool in[maxl];
 16 
 17 inline void add(int u,int v)
 18 {
 19     e[++cnt].to=v;e[cnt].nxt=ehead[u];ehead[u]=cnt;
 20 }
 21 
 22 inline void prework()
 23 {
 24     for(int i=1;i<=n;i++)
 25         ehead[i]=0,f[i]=0,in[i]=false;
 26     int u,v;
 27     cnt=0;
 28     for(int i=1;i<=n-1;i++)
 29     {
 30         scanf("%d%d",&u,&v);
 31         add(u,v);add(v,u);
 32     }
 33     for(int i=1;i<=k;i++)
 34     {
 35         scanf("%d",&u);
 36         in[u]=true;
 37     }
 38 }
 39 
 40 inline void gets(int u,int fa)
 41 {
 42     int v,tmp;
 43     if(in[u])
 44         mxflag[u]=true;
 45     for(int i=ehead[u];i;i=e[i].nxt)
 46     {
 47         v=e[i].to;
 48         if(v==fa) continue;
 49         gets(v,u);
 50         if(in[v])
 51         {
 52             in[u]=true;
 53             if(mxflag[v])
 54             {
 55                 tmp=mxs[v]+1;
 56                 if(tmp>mxs[u])
 57                 {
 58                     secmxs[u]=mxs[u],mxs[u]=tmp;
 59                     secmxflag[u]=mxflag[u];
 60                     mxflag[u]=true;
 61                 }
 62                 else if(tmp>secmxs[u])
 63                     secmxs[u]=tmp,secmxflag[u]=true;
 64             }
 65         }
 66     }    
 67 }
 68 
 69 inline void getf(int u,int fa)
 70 {
 71     int v,tmp;
 72     if(u==1) f[1]=mxs[1];
 73     else
 74     {
 75         if(mxs[u]+1==mxs[fa])
 76         {
 77             if(secmxflag[fa])
 78             {
 79                 mxfa[u]=max(mxfa[u],secmxs[fa]+1);
 80                 faflag[u]=true;
 81             }
 82         }
 83         else
 84         {
 85             if(mxflag[fa])
 86             {
 87                 mxfa[u]=max(mxfa[u],mxs[fa]+1);
 88                 faflag[u]=true;
 89             }
 90         }
 91         if(faflag[fa])
 92         {    
 93             mxfa[u]=max(mxfa[u],mxfa[fa]+1);
 94             faflag[u]=true;
 95         }
 96     }
 97     for(int i=ehead[u];i;i=e[i].nxt)
 98     {
 99         v=e[i].to;
100         if(v==fa) continue;
101         getf(v,u);
102     }
103     if(faflag[u])
104         f[u]=mxfa[u];
105     if(mxs[u])
106         f[u]=max(f[u],mxs[u]);
107 }
108 
109 inline void mainwork()
110 {
111     for(int i=1;i<=n;i++){
112         f[i]=0,mxs[i]=0,secmxs[i]=0,mxfa[i]=0;
113         mxflag[i]=secmxflag[i]=faflag[i]=false;
114     }
115     gets(1,0);
116     mxfa[1]=0;
117     getf(1,0);
118     ans=inf;
119     for(int i=1;i<=n;i++)
120         ans=min(ans,f[i]);
121 }
122 
123 inline void print()
124 {
125     printf("%lld
",ans);
126 }
127 
128 int main()
129 {
130     while(~scanf("%d%d",&n,&k))
131     {
132         prework();
133         mainwork();
134         print();
135     }
136     return 0;
137 }

View Code

B xor

题解：https://blog.csdn.net/liufengwei1/article/details/97620377

  1 #include<bits/stdc++.h>
  2 #define maxl 50010
  3 using namespace std;
  4  
  5 int n,m,q;
  6 int sz[maxl];
  7 unsigned int a[maxl][33];
  8 struct LB
  9 {
 10     int cnt;bool flag;
 11     unsigned int p[32],f[32];
 12     inline void init()
 13     {
 14         cnt=0;flag=false;
 15         memset(f,0,sizeof(f));
 16         memset(p,0,sizeof(p));
 17     }
 18     inline bool insert(unsigned int x)
 19     {
 20         for(int i=31;i>=0;i--)
 21         if(x&(1ll<<i))
 22         {
 23             if(!p[i])
 24             {
 25                 p[i]=x;
 26                 break;
 27             }
 28             x^=p[i];
 29         }
 30         return x>0;
 31     }
 32     inline void insert2(unsigned int x)
 33     {
 34         unsigned int tmp=x;
 35         for(int i=31;i>=0;i--)
 36         if(x>>i)
 37         {
 38             if(!p[i])
 39             {
 40                 f[i]=tmp;p[i]=x;
 41                 return;
 42             }
 43             x^=p[i];tmp^=f[i];
 44         }
 45     }
 46     inline bool find(unsigned int x)
 47     {
 48         for(int i=31;i>=0;i--)
 49         if(x>>i)
 50         {
 51             if(!p[i]) return 0;
 52             x^=p[i];
 53         }
 54         return x==0;
 55     }
 56     unsigned int calc(unsigned int x)
 57     {
 58         unsigned int ret=0;
 59         for(int i=31;i>=0;i--)
 60         if(x>>i)
 61         {
 62             ret^=f[i];
 63             x^=p[i];
 64         }
 65         return ret;    
 66     } 
 67 };
 68 struct node
 69 {
 70     int l,r;
 71     LB xxj;
 72 }tree[maxl*4];
 73  
 74 inline void pushup(int k)
 75 {
 76     LB tmp=tree[k<<1].xxj;
 77     LB ans;
 78     ans.init();
 79     for(int i=31;i>=0;i--)
 80     {
 81         unsigned int x=tree[k<<1|1].xxj.p[i];
 82         if(tmp.find(x))
 83             ans.insert(x^tmp.calc(x));
 84         else
 85             tmp.insert2(x);
 86     }
 87     tree[k].xxj=ans;
 88 }
 89  
 90 inline void build(int k,int l,int r)
 91 {
 92     tree[k].l=l;tree[k].r=r;tree[k].xxj.init();
 93     if(l==r)
 94     {
 95         for(int i=1;i<=sz[l];i++)
 96             tree[k].xxj.insert(a[l][i]);
 97         return;
 98     }
 99     int mid=(l+r)>>1;
100     build(k<<1,l,mid);
101     build(k<<1|1,mid+1,r);
102     pushup(k);
103 }
104  
105 inline void prework()
106 {
107     for(int i=1;i<=n;i++)
108     {
109         scanf("%d",&sz[i]);
110         for(int j=1;j<=sz[i];j++)
111             scanf("%u",&a[i][j]);
112     }
113     build(1,1,n);
114 }
115  
116 inline bool query(int k,int l,int r,unsigned int x)
117 {
118     if(tree[k].l==l && tree[k].r==r)
119         return tree[k].xxj.find(x);
120     int mid=(tree[k].l+tree[k].r)>>1;
121     if(r<=mid) 
122         return query(k<<1,l,r,x);
123     else if(l>mid)
124         return query(k<<1|1,l,r,x);
125     else
126         return query(k<<1,l,mid,x) && query(k<<1|1,mid+1,r,x);
127 }
128  
129 inline void mainwork()
130 {
131     int l,r;unsigned int x;
132     for(int i=1;i<=q;i++)
133     {
134         scanf("%d%d%u",&l,&r,&x);
135         if(query(1,l,r,x))
136             puts("YES");
137         else
138             puts("NO");
139     }
140 }
141  
142 int main()
143 {
144     while(~scanf("%d%d",&n,&q))
145     {
146         prework();
147         mainwork();
148     //    print();
149     }
150     return 0;
151 }

View Code

C sequence

单调栈维护每个数字的最小值统治的区域，线段树找区间最值

  1 #include <bits/stdc++.h>
  2 #define my_max(a, b) ((a) > (b) ? (a) : (b))
  3 #define my_min(a, b) ((a) < (b) ? (a) : (b))
  4 #define fi first
  5 #define se second
  6 #define pb push_back
  7 #define eb emplace_back
  8 #define rep(i, s, t) for(int i = (int)(s); i <= (int)(t); i++)
  9 #define rev(i, t, s) for(int i = (int)(t); i >= (int)(s); i--)
 10 #define lson rt << 1
 11 #define rson rt << 1 | 1
 12 #define sz(x) (int)(x).size()
 13   
 14 typedef long long ll;
 15 typedef long double lb;
 16 typedef std::pair<int, int> pii;
 17 typedef std::pair<ll, ll> pll;
 18 typedef std::vector<int> VI;
 19 typedef std::vector<ll> VL;
 20 #include <ext/pb_ds/tree_policy.hpp>
 21 #include <ext/pb_ds/assoc_container.hpp>
 22 using namespace __gnu_pbds;
 23 using namespace std;
 24 typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update> rbtree;
 25 ll gcd(ll x, ll y){ return y % x == 0 ? x : gcd(y % x, x); }
 26 template<class T>T my_abs(T a){ if(a < 0) a = -a; return a; }
 27 inline ll read()
 28 {
 29     ll ret = 0, sign = 1;
 30     char c = getchar();
 31     while(c < '0' || c > '9'){ if(c == '-') sign = -1; c = getchar();}
 32     while(c >= '0' && c <= '9'){ ret = ret * 10 + c - '0'; c = getchar(); }
 33     return ret * sign;
 34 }
 35 inline void write(ll x)
 36 {
 37     if(x < 0){
 38         putchar('-');
 39         x = -x;
 40     }
 41     if(x > 9) write(x / 10);
 42     putchar(x % 10 + '0');
 43 }
 44   
 45 const int maxn = 5e6 + 7;
 46 const ll inf = 1e18;
 47 int n;
 48 struct SegmentTree{
 49     ll Max[maxn << 2], Min[maxn << 2];
 50     void init(ll *a){
 51         build(1, 1, n, a);
 52     }
 53     void push_up(int rt)
 54     {
 55         Min[rt] = min(Min[lson], Min[rson]);
 56         Max[rt] = max(Max[lson], Max[rson]);
 57     }
 58     void build(int rt, int l, int r, ll *a)
 59     {
 60         if(l == r){
 61             Max[rt] = Min[rt] = a[l];
 62             return ;
 63         }
 64         int mid = (l + r) >> 1;
 65         build(lson, l, mid, a);
 66         build(rson, mid + 1, r, a);
 67         push_up(rt);
 68     }
 69     ll queryMin(int ql, int qr, int rt = 1, int l = 1, int r = n){
 70         if(ql == l && qr == r) return Min[rt];
 71         int mid = (l + r) >> 1;
 72         ll res = inf;
 73         if(qr <= mid) res = queryMin(ql, qr, lson, l, mid);
 74         else if(ql > mid) res = queryMin(ql, qr, rson, mid + 1, r);
 75         else{
 76             res = min(queryMin(ql, mid, lson, l, mid), queryMin(mid + 1, qr, rson, mid + 1, r));
 77         }
 78         return res;
 79     }
 80     ll queryMax(int ql, int qr, int rt = 1, int l = 1, int r = n)
 81     {
 82         if(ql == l && qr == r) return Max[rt];
 83         int mid = (l + r) >> 1;
 84         ll res = -inf;
 85         if(qr <= mid) res = queryMax(ql, qr, lson, l, mid);
 86         else if(ql > mid) res = queryMax(ql, qr, rson, mid + 1, r);
 87         else{
 88             res = max(queryMax(ql, mid, lson, l, mid), queryMax(mid + 1, qr, rson, mid + 1, r));
 89         }
 90         return res;
 91     }
 92 }pre;
 93 ll presum[maxn], a[maxn],b[maxn];
 94 int L[maxn], R[maxn], st[maxn], top;
 95 int main()
 96 {
 97  
 98     scanf("%d", &n);
 99     presum[1] = 0;
100      
101     for(int i = 2; i <= n + 1; i++)
102         scanf("%lld", a + i);
103     for(int i = 2; i <= n + 1; i++)
104     {
105         scanf("%lld", b + i);
106         presum[i] = presum[i - 1] + b[i];
107     }
108     top = 0;
109     for(int i = 2; i <= n + 1; i++)
110     {
111         while(top > 0 && a[st[top]] >= a[i]) top--;
112         if(top > 0) L[i] = st[top] + 1;
113         else L[i] = 1;
114         st[++top] = i;
115     }
116     top = 0;
117     for(int i = n + 1; i >= 2; i--)
118     {
119         while(top > 0 && a[st[top]] >= a[i]) top--;
120         if(top > 0) R[i] = st[top] - 1;
121         else R[i] = n + 1;
122         st[++top] = i;
123     }
124     n++;
125     pre.init(presum);
126     ll ans = -inf;
127      
128     for(int i = 2; i <= n; i++)
129     {
130         int l = L[i], r = R[i];
131         l = max(l - 1, 1);
132        
133         ll Maxl = pre.queryMax(l, i), Minl = pre.queryMin(l, i);
134         ll Maxr = pre.queryMax(i, r), Minr = pre.queryMin(i, r);
135         ll ans1 = (Maxr - Minl) * a[i];
136         ll ans2 = (Minr - Maxl) * a[i];
137          
138         ans1 = max(ans1, ans2); ans1 = max(ans1, a[i] * a[i]);
139         ans = max(ans, ans1);
140     }
141     printf("%lld
", ans);
142     return 0;
143  
144 }

View Code

D triples I

题解：https://blog.csdn.net/liufengwei1/article/details/97551038

  1 #include<bits/stdc++.h>
  2 #define maxl 65
  3 using namespace std;
  4  
  5 long long top,anscnt;
  6 long long a;
  7 long long anum[maxl];
  8 long long ans[3],mi[maxl];
  9 bool in[maxl];
 10  
 11 inline void prework()
 12 {
 13     scanf("%lld",&a);
 14     long long x=a;
 15     top=0;long long cnt=0;
 16     while(x>0)
 17     {
 18         cnt++;
 19         if(x%2ll==1ll)
 20             anum[++top]=cnt;
 21         x/=2ll;
 22     }
 23 }
 24  
 25 inline void mainwork()
 26 {
 27     long long x;
 28     if(a%3ll==0ll)
 29     {
 30         anscnt=1;
 31         ans[1]=a;
 32         return;
 33     }
 34     long long cnt1=0,cnt2=0,sum=0,num1,num2;
 35     for(long long i=1;i<=top;i++)
 36     {
 37         if(anum[i]%2ll==1ll)
 38             cnt1++;
 39         else
 40             cnt2++;
 41     }
 42     sum=cnt1+2*cnt2;
 43     if(sum%3ll==1ll)
 44     {
 45         if(cnt1>=1ll)
 46         {
 47             num1=cnt1-1;
 48             num2=cnt2;
 49         }
 50         else
 51         {
 52             num1=cnt1;
 53             num2=cnt2-2;
 54         }
 55     }
 56     else
 57     {
 58         if(cnt2>=1ll)
 59         {
 60             num1=cnt1;
 61             num2=cnt2-1;
 62         }
 63         else
 64         {
 65             num1=cnt1-2;
 66             num2=cnt2;
 67         }
 68     }
 69     long long rescnt1=cnt1-num1,rescnt2=cnt2-num2;
 70     anscnt=2;
 71     ans[1]=0;ans[2]=0;
 72     cnt1=0;cnt2=0;
 73     for(long long i=1;i<=top;i++)
 74         in[i]=false;
 75     for(long long i=1;i<=top;i++)
 76     {
 77         if(anum[i]%2ll==1ll)
 78         {
 79             if(cnt1<num1)
 80                 cnt1++,ans[1]|=mi[anum[i]],in[i]=true;
 81             else
 82                 ans[2]|=mi[anum[i]];
 83         }
 84         else
 85         {
 86             if(cnt2<num2)
 87                 cnt2++,ans[1]|=mi[anum[i]],in[i]=true;
 88             else
 89                 ans[2]|=mi[anum[i]];
 90         }
 91     }
 92     if((rescnt1+2*rescnt2)%3ll==1)
 93     {
 94         if(cnt2>0)
 95         {
 96             for(long long i=1;i<=top;i++)
 97             if(in[i] && anum[i]%2==0)
 98             {
 99                 ans[2]|=mi[anum[i]];
100                 break;
101             }
102         }
103         else
104         {
105             bool flag=false;
106             for(long long i=1;i<=top;i++)
107             if(in[i] && anum[i]%2==1)
108             {
109                 ans[2]|=mi[anum[i]];
110                 if(flag) break;
111                 flag=true;
112             }
113         }
114     }
115     else
116     {
117         if(cnt1>0)
118         {
119             for(long long i=1;i<=top;i++)
120             if(in[i] && anum[i]%2==1)
121             {
122                 ans[2]|=mi[anum[i]];
123                 break;
124             }
125         }
126         else
127         {
128             bool flag=false;
129             for(long long i=1;i<=top;i++)
130             if(in[i] && anum[i]%2==0)
131             {
132                 ans[2]|=mi[anum[i]];
133                 if(flag) break;
134                 flag=true;
135             }
136         }
137     }
138 }
139  
140 inline void print()
141 {
142     printf("%lld ",anscnt);
143     for(long long i=1;i<=anscnt;i++)
144         printf("%lld%c",ans[i],(i==anscnt)?'
':' ');
145 }
146  
147 int main()
148 {
149     mi[1]=1;
150     for(long long i=2;i<=63;i++)
151         mi[i]=mi[i-1]*2ll;
152     long long t;
153     scanf("%lld",&t);
154     for(long long i=1;i<=t;i++)
155     {
156         prework();
157         mainwork();
158         print();
159     }
160     return 0;
161 }

View Code

E triples II

unsolved

F merge

unsolved

G tree

unsolved

H RNGs

unsolved

I string

题解：https://www.cnblogs.com/songorz/p/11257819.html

  1 #include<bits/stdc++.h>
  2 using namespace std;
  3 typedef long long ll;
  4 #define pii pair<int,int>
  5 #define pil pair<int,long long>
  6 const int INF=0x3f3f3f3f;
  7 const ll inf=0x3f3f3f3f3f3f3f3fll;
  8 inline int read()
  9 {
 10     int x=0,f=1;char ch=getchar();
 11     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
 12     while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}    
 13     return x*f;
 14 }
 15 const int maxn=4e5+10;
 16 const int MAXN=4e5+10; 
 17 char str[maxn];
 18 int s[maxn];
 19 ll ans;
 20 struct SAM{
 21     int l[maxn<<1],fa[maxn<<1],nxt[maxn<<1][30];
 22     int last,cnt;
 23     
 24     void Init()
 25     {
 26         ans=0;last=cnt=1;
 27         l[cnt]=fa[cnt]=0;
 28         memset(nxt[cnt],0,sizeof(nxt[cnt]));        
 29     }
 30     
 31     int NewNode()
 32     {
 33         ++cnt;
 34         memset(nxt[cnt],0,sizeof(nxt[cnt]));
 35         l[cnt]=fa[cnt]=0;
 36         return cnt;    
 37     }
 38     
 39     void Insert(int ch)
 40     {
 41         int np=NewNode(),p=last;
 42         last=np; l[np]=l[p]+1;
 43         while(p&&!nxt[p][ch]) nxt[p][ch]=np,p=fa[p];
 44         if(!p) fa[np]=1;
 45         else
 46         {
 47             int q=nxt[p][ch];
 48             if(l[p]+1==l[q]) fa[np]=q;
 49             else
 50             {
 51                 int nq=NewNode();
 52                 memcpy(nxt[nq],nxt[q],sizeof(nxt[q]));
 53                 fa[nq]=fa[q];
 54                 l[nq]=l[p]+1;
 55                 fa[np]=fa[q]=nq;
 56                 while(nxt[p][ch]==q) nxt[p][ch]=nq,p=fa[p];
 57             }    
 58         }    
 59         ans+=1ll*(l[last]-l[fa[last]]);    
 60     }
 61     
 62 }sam;
 63 
 64 struct Palindromic_Tree{
 65     int next[MAXN][26];
 66     int fail[MAXN];
 67     int cnt[MAXN];
 68     int num[MAXN];
 69     int len[MAXN];
 70     int S[MAXN];
 71     int last;
 72     int n;
 73     int p;
 74  
 75     int newnode(int l) 
 76     {
 77         for(int i=0;i<26;++i) next[p][i]=0;
 78         cnt[p]=0;
 79         num[p]=0;
 80         len[p]=l;
 81         return p++;
 82     }
 83  
 84     void Init() 
 85     {
 86         p=0;
 87         newnode( 0);
 88         newnode(-1);
 89         last=0;
 90         n=0;
 91         S[n]=-1;
 92         fail[0]=1;
 93     }
 94  
 95     int get_fail(int x)
 96     {
 97         while(S[n-len[x]-1]!=S[n])x=fail[x] ;
 98         return x ;
 99     }
100  
101     void add(int c) 
102     {
103         S[++ n]=c;
104         int cur=get_fail(last) ;
105         if(!next[cur][c]) 
106         {
107             int now=newnode(len[cur]+2) ;
108             fail[now]=next[get_fail(fail[cur])][c] ;
109             next[cur][c]=now ;
110             num[now]=num[fail[now]]+1;
111         }
112         last=next[cur][c];
113         cnt[last]++;
114     }
115  
116     ll count() 
117     {
118         ll res=p*1ll;
119         for(int i=p-1;i>=0;--i) cnt[fail[i]]+=cnt[i];
120         //for(int i=1;i<=p;++i) res+=cnt[i];
121         //cout<<"res "<<res<<endl;
122         return (res-2);
123     }
124 } pam;
125 
126 int main()
127 {
128     scanf("%s",str);
129     int len=strlen(str);
130     
131     sam.Init();
132     for(int i=0;i<len;++i) sam.Insert(str[i]-'a');
133     sam.Insert(28);
134     for(int i=len-1;i>=0;--i) sam.Insert(str[i]-'a');
135     ans-=1ll*(len+1)*(len+1);
136     //cout<<"ans "<<ans<<endl;
137     pam.Init();
138     for(int i=0;i<len;++i) pam.add(str[i]-'a');
139     ans=ans+pam.count();
140     
141     printf("%lld
",(ans/2ll));
142     
143     
144     return 0;    
145 }

View Code

J free

k维最短路

  1 #include<bits/stdc++.h>
  2 #define maxl 1010
  3 using namespace std;
  4 
  5 const int inf=2e9;
  6 
  7 int n,m,S,T,K,cnt;
  8 int ehead[maxl];
  9 int dis[maxl][maxl];
 10 struct ed
 11 {
 12     int to,nxt,l;
 13 }e[maxl*2];
 14 struct node
 15 {
 16     int val,k,u;
 17     bool operator > (const node &b)const
 18     {
 19         if(val==b.val)
 20         {
 21             if(k==b.k)
 22                 return u>b.u;
 23             else
 24                 return k>b.k;
 25         }
 26         else return val>b.val;
 27     }
 28 };
 29 priority_queue<node,vector<node>,greater<node>> q;
 30 
 31 inline void add(int u,int v,int l)
 32 {
 33     e[++cnt].to=v;e[cnt].l=l;
 34     e[cnt].nxt=ehead[u];ehead[u]=cnt;
 35 }
 36 
 37 inline void prework()
 38 {
 39     for(int i=1;i<=n;i++)
 40         ehead[i]=0;
 41     scanf("%d%d%d",&S,&T,&K);
 42     cnt=0;int u,v,l;
 43     for(int i=1;i<=m;i++)
 44     {
 45         scanf("%d%d%d",&u,&v,&l);
 46         if(u!=v)
 47             add(u,v,l),add(v,u,l);
 48     }
 49 }
 50 
 51 inline void mainwork()
 52 {
 53     for(int i=1;i<=n;i++)
 54         for(int k=0;k<=K;k++)
 55             dis[i][k]=inf;
 56     node d;int u,v;
 57     dis[S][0]=0;
 58     q.push(node{0,0,S});
 59     while(!q.empty())
 60     {
 61         do
 62         {
 63             d=q.top();q.pop();
 64             u=d.u;
 65         }
 66         while(d.val>dis[u][d.k] && !q.empty());
 67         for(int i=ehead[u];i;i=e[i].nxt)
 68         {
 69             v=e[i].to;
 70             if(d.k<K)
 71             {
 72                 if(d.val<dis[v][d.k+1])
 73                 {
 74                     dis[v][d.k+1]=d.val;
 75                     q.push(node{d.val,d.k+1,v});
 76                 }
 77             }
 78             if(d.val+e[i].l<dis[v][d.k])
 79             {
 80                 dis[v][d.k]=d.val+e[i].l;
 81                 q.push(node{dis[v][d.k],d.k,v});
 82             }
 83         }
 84     }
 85 }
 86 
 87 inline void print()
 88 {
 89     int ans=inf;
 90     for(int k=0;k<=K;k++)
 91         ans=min(dis[T][k],ans);
 92     printf("%d
",ans);
 93 }
 94 
 95 int main()
 96 {
 97     while(~scanf("%d%d",&n,&m))
 98     {
 99         prework();
100         mainwork();
101         print();
102     }
103     return 0;
104 }

View Code

K number

DP计数

 1 #include<bits/stdc++.h>
 2 #define maxl 100010
 3 using namespace std;
 4 
 5 int n;
 6 int sum[4];
 7 char s[maxl];
 8 long long ans;
 9 
10 inline void prework()
11 {
12     n=strlen(s+1);
13     for(int i=0;i<=3;i++)
14         sum[i]=0;
15 }
16 
17 inline void mainwork()
18 {
19     ans=0;
20     int cnt1=0,cnt2=0,tmp=0;
21     sum[0]=1;
22     for(int i=1;i<=n;i++)
23     {
24         tmp=(tmp+s[i]-'0')%3;
25         if(s[i]=='0' && s[i-1]=='0')
26             ans+=sum[tmp];
27         else if(s[i]=='0')
28             ans+=1;
29         sum[tmp]+=1;
30     }
31 }
32 
33 inline void print()
34 {
35     printf("%lld
",ans);
36 }
37 
38 int main()
39 {
40     while(~scanf("%s",s+1))
41     {
42         prework();
43         mainwork();
44         print();
45     }
46     return 0;
47 }

View Code

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原文地址：https://www.cnblogs.com/csushl/p/11280412.html