还有13天NOI，把各种乱七八糟的算法都重新过一遍还是比较有必要的。。。
//HDU 5046 Airport //DancingLink #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; #define MAXN 110 #define MAXD MAXN*MAXN #define INF 0x3f3f3f3f #define LL "%lld" typedef long long qword; struct point { qword x,y; }; point pl[MAXN]; qword dis(point p1,point p2) { return abs(p1.x-p2.x)+abs(p1.y-p2.y); } int L[MAXD],R[MAXD],U[MAXD],D[MAXD]; int ptr[MAXN]; int tot[MAXN]; int col[MAXD]; int topd=0; int head=0; void cover(int now) { for (int i=R[now];i!=now;i=R[i]) { for (int j=U[i];j!=i;j=U[j]) { L[R[j]]=L[j]; R[L[j]]=R[j]; } } for (int j=U[now];j!=now;j=U[j]) { L[R[j]]=L[j]; R[L[j]]=R[j]; } } void recover(int now) { for (int i=R[now];i!=now;i=R[i]) for (int j=U[i];j!=i;j=U[j]) L[R[j]]=R[L[j]]=j; for (int j=U[now];j!=now;j=U[j]) L[R[j]]=R[L[j]]=j; } int vv[MAXN]; int ff() { int ret=0; for (int i=R[head];i!=head;i=R[i]) vv[col[i]]=true; for (int i=R[head];i!=head;i=R[i]) { if (vv[col[i]]) { ret++; for (int j=D[i];j!=i;j=D[j]) { for (int k=R[j];k!=j;k=R[k]) { vv[col[k]]=false; } } } } return ret; } bool solve(int trst) { if (L[head]==R[head] && L[head]==head)return true; if (!trst)return false; if (trst<ff())return false; pair<int,int> mnv=make_pair(INF,0); for (int i=R[head];i!=head;i=R[i]) mnv=min(mnv,make_pair(tot[i],i)); int now=mnv.second; for (int i=D[now];i!=now;i=D[i]) { cover(i); if (solve(trst-1))return true; recover(i); } return false; } int main() { freopen("input.txt","r",stdin); int nn; int n,m; int x,y,z; scanf("%d",&nn); int caseid=0; while (nn--) { caseid++; scanf("%d%d",&n,&m); for (int i=1;i<=n;i++) scanf(LL LL ,&pl[i].x,&pl[i].y); memset(ptr,0,sizeof(ptr[0])*(n+10)); memset(tot,0,sizeof(tot[0])*(n+10)); qword l=-1,r=1ll<<33,mid; while (l+1<r) { mid=(l+r)>>1; topd=0; head=++topd; L[head]=R[head]=head; for (int i=1;i<=n;i++) { int np=++topd; col[np]=i; R[np]=head; L[np]=L[head]; L[R[np]]=np; R[L[np]]=np; D[np]=U[np]=np; ptr[i]=np; } for (int i=1;i<=n;i++) { int last=0; for (int j=1;j<=n;j++) { if (dis(pl[i],pl[j])<=mid) { int np=++topd; col[np]=j; tot[ptr[j]]++; D[np]=ptr[j]; U[np]=U[ptr[j]]; D[U[np]]=U[D[np]]=np; if (!last) { L[np]=R[np]=np; }else { L[np]=last; R[np]=R[last]; L[R[np]]=R[L[np]]=np; } last=np; } } } if (solve(m)) { r=mid; }else { l=mid; } } printf("Case #%d: "LL" ",caseid,r); } }
半年没写DLX了，还能在30分钟内写出来，感觉不错。
DLX分为精确覆盖和完全覆盖，两者的区别大概是在cover和recover之中。
另外DLX也是需要剪枝的。
//bzoj 1135 #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; #define MAXN 210000 #define MAXT MAXN*5 #define lch (now<<1) #define rch (now<<1^1) #define smid ((l+r)>>1) typedef long long qword; struct sgt_node { int lc,rc; qword lx,rx,mx; qword sum; }sgt[MAXT]; void update(int now) { sgt[now].lx=max(sgt[lch].lx,sgt[lch].sum+sgt[rch].lx); sgt[now].rx=max(sgt[rch].rx,sgt[rch].sum+sgt[lch].rx); sgt[now].sum=sgt[lch].sum+sgt[rch].sum; sgt[now].mx=max(max(sgt[lch].mx,sgt[rch].mx),sgt[lch].rx+sgt[rch].lx); } void Build_sgt(int now,int l,int r,int v) { if (l==r) { sgt[now].sum=v; sgt[now].lx=sgt[now].rx=sgt[now].mx=v; return ; } Build_sgt(lch,l,smid,v); Build_sgt(rch,smid+1,r,v); update(now); } void Modify_sgt(int now,int l,int r,int pos,int v) { if (l==r) { sgt[now].sum+=v; sgt[now].lx+=v; sgt[now].rx+=v; sgt[now].mx+=v; return ; } if (pos<=smid) Modify_sgt(lch,l,smid,pos,v); else Modify_sgt(rch,smid+1,r,pos,v); update(now); } int main() { freopen("input.txt","r",stdin); int n,m,t,d; int x,y; scanf("%d%d%d%d",&n,&m,&t,&d); Build_sgt(1,1,n,-t); for (int i=0;i<m;i++) { scanf("%d%d",&x,&y); Modify_sgt(1,1,n,x,y); if (sgt[1].mx>(qword)t*d) { printf("NIE "); }else { printf("TAK "); } }
hall定理用于二分图匹配相关问题，在要求方案时用贪心或匈牙利算法，运用hall定理有利于优化时间复杂度。
//bzoj 3270 #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cassert> #include<cmath> using namespace std; #define MAXN 1010 #define MAXV MAXN #define MAXE MAXN*20 typedef double real; const real eps = 1e-7; struct Edge { int np; Edge *next; }E[MAXE],*V[MAXV]; int tope=-1; void addedge(int x,int y) { E[++tope].np=y; E[tope].next=V[x]; V[x]=&E[tope]; } real ps[MAXN]; real mat[MAXN][MAXN]; real res[MAXN]; int deg[MAXN]; int main() { freopen("input.txt","r",stdin); int n,m; int a,b; int x,y,z; scanf("%d%d",&n,&m); scanf("%d%d",&a,&b); a--;b--; for (int i=0;i<m;i++) { scanf("%d%d",&x,&y); x--;y--; addedge(x,y); addedge(y,x); deg[x]++; deg[y]++; } for (int i=0;i<n;i++) scanf("%lf",&ps[i]); for (int i=0;i<n;i++) { for (int j=0;j<n;j++) { if (i==j) { mat[i*n+j][i*n+j]=1; continue; } Edge *ne1,*ne2; for (ne1=V[i];ne1;ne1=ne1->next) for (ne2=V[j];ne2;ne2=ne2->next) mat[ne1->np*n+ne2->np][i*n+j]=-1*(1-ps[i])*(1-ps[j])/deg[i]/deg[j]; Edge *ne; for (ne=V[i];ne;ne=ne->next) mat[ne->np*n+j][i*n+j]=-1*ps[j]*(1-ps[i])/deg[i]; for (ne=V[j];ne;ne=ne->next) mat[i*n+ne->np][i*n+j]=-1*(1-ps[j])*ps[i]/deg[j]; mat[i*n+j][i*n+j]=1-ps[i]*ps[j]; mat[i*n+j][n*n]=0; } } mat[a*n+b][n*n]=1; int l=n*n; for (int i=0;i<l;i++) { int x=-1; for (int j=i;j<=l;j++) { if (abs(mat[j][i])>eps) x=j; } assert(x!=-1); if (x!=i) for(int j=0;j<=l;j++) swap(mat[i][j],mat[x][j]); for (int j=i+1;j<l;j++) { real tmp=mat[j][i]/mat[i][i]; for (int k=i;k<=l;k++) { mat[j][k]-=mat[i][k]*tmp; } } } for (int i=l-1;i>=0;i--) { real tmp=mat[i][n*n]; for (int j=i+1;j<n*n;j++) tmp-=res[j]*mat[i][j]; res[i]=tmp/mat[i][i]; } for (int i=0;i<n;i++) printf("%.6lf ",res[i*n+i]); }
概率期望dp以及高消在很多时候都是可以转化的，而适当的转化会使思维难度大大降低。
//Miller Rabin //HDU 2138 #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; typedef long long qword; qword pow_mod(qword x,qword y,int mod) { qword ret=1%mod; while (y) { if (y&1)ret=ret*x%mod; x=x*x%mod; y>>=1; } return ret; } bool MillerRabin(qword a,int n) { int x=n-1,y=0; while ((x&1)==0)x>>=1,y++; a=pow_mod(a,x,n); int pe=a; for (int i=0;i<y;i++) { pe=a; a=a*a%n; if (a==1 && pe!=n-1 && pe!=1)return false; } return a==1?true:false; } int main() { // freopen("input.txt","r",stdin); int n; int x; int s[6]={2,3,5,7,17,61}; while (~scanf("%d",&n)) { int ans=0; for (int i=0;i<n;i++) { scanf("%d",&x); if (x==1) { continue; }else if (x==2) { ans++; continue; } bool flag=true; for (int j=0;j<6;j++) { if (s[j]>=x)continue; if (!MillerRabin(s[j],x)) { flag=false; break; } } if (flag) { ans++; } } printf("%d ",ans);
这种时候还理解什么，赶快背啊！
#include<iostream> #include<cstring> #include<algorithm> #include<cstdio> using namespace std; #define MAXN 110000 #define MAXM 210 #define BIG 1000000000000000LL typedef long long qword; int a[MAXN]; qword s[MAXN]; qword dp[2][MAXN]; struct point { qword x,y; }; qword area(point p1,point p2,point p3) { return (p1.x-p2.x)*(p1.y-p3.y) - (p1.y-p2.y)*(p1.x-p3.x); } point q[MAXN]; int main() { freopen("input.txt","r",stdin); freopen("output.txt","w",stdout); int n,m; scanf("%d%d",&n,&m); for (int i=1;i<=n;i++) { scanf("%d",&a[i]); //if (!a[i])i--,n--; } for (int i=1;i<=n;i++) s[i]=s[i-1]+a[i]; for (int i=0;i<2;i++) for (int j=0;j<=n;j++) dp[i][j]=-BIG; int head,tail; point pt; dp[0][0]=0; for (int i=1;i<=m;i++) { head=0; tail=-1; for (int j=1;j<=n;j++) { pt.x=s[j-1]; pt.y=-s[n]*s[j-1]+dp[(i&1)^1][j-1]; while (head<tail && area(q[tail-1],q[tail],pt)>=0) { if (area(q[tail-1],q[tail],pt)==0) { cout<<"haha"<<endl; } tail--; } q[++tail]=pt; while (head<tail && s[j]*q[head].x+q[head].y<=s[j]*q[head+1].x+q[head+1].y) head++; dp[i&1][j]=s[j]*q[head].x+q[head].y + s[n]*s[j]-s[j]*s[j]; // printf("%lld ",dp[i&1][j]); } dp[i&1][0]=-BIG; // printf(" "); } qword ans=0; for (int i=1;i<=n;i++) ans=max(ans,dp[m&1][i]); printf("%lld ",ans); }
//POJ 1160 #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; #define MAXN 3100 #define MAXM 40 #define INF 0x3f3f3f3f int w[MAXN][MAXN],w1[MAXN][MAXN]; int a[MAXN]; int dp[MAXM][MAXN]; pair<int,int> srange[MAXN]; int stack[MAXN]; int tops=-1; int main() { freopen("input.txt","r",stdin); int n,m; int x,y,z; scanf("%d%d",&n,&m); for (int i=1;i<=n;i++) scanf("%d",a+i); for (int i=1;i<=n;i++) { for (int j=i;j<=n;j++) { if ((i+j)/2==(i+j-1)/2) w[i][j]=w[i][j-1]+a[j]-a[(i+j)/2]; else w[i][j]=w[i][j-1]+(a[(i+j)/2]-a[(i+j-1)/2])*(((i+j)/2-i) - (j-(i+j)/2)) +a[j]-a[(i+j)/2]; } } memset(dp,INF,sizeof(dp)); dp[0][0]=0; srange[++tops]=make_pair(1,n); stack[tops]=0; for (int i=1;i<=m;i++) { while (~tops) { for (int j=srange[tops].first;j<=srange[tops].second;j++) dp[i][j]=dp[i-1][stack[tops]]+w[stack[tops]+1][j]; tops--; } for (int j=1;j<=n;j++) { while (~tops && dp[i][j]+w[j+1][srange[tops].first]<dp[i][stack[tops]]+w[stack[tops]+1][srange[tops].first]) tops--; if (~tops) { int l=srange[tops].first; int r=srange[tops].second+1; while (l+1<r) { int mid=(l+r)>>1; if (dp[i][j]+w[j+1][mid]<dp[i][stack[tops]]+w[stack[tops]+1][mid]) r=mid; else l=mid; } srange[tops].second=l; if (r<=n) { srange[++tops]=make_pair(r,n); stack[tops]=j; } }else { srange[++tops]=make_pair(1,n); stack[tops]=j; } } } int ans=dp[m][n]; printf("%d ",ans); return 0; }
//bzoj 1856 #include<iostream> #include<cstring> #include<cstdio> #include<algorithm> using namespace std; #define MOD 20100403 #define MAXN 2001000 typedef long long qword; qword fact[MAXN]; qword pow_mod(qword x,qword y) { qword ret=1; while (y) { if (y&1)ret=ret*x%MOD; x=x*x%MOD; y>>=1; } return ret; } qword C(int x,int y) { if (x<y)return 0; return fact[x]*pow_mod(fact[x-y],MOD-2)%MOD*pow_mod(fact[y],MOD-2)%MOD; } int main() { //freopen("input.txt","r",stdin); int n,m; scanf("%d%d",&n,&m); fact[0]=1; for (int i=1;i<=n+m;i++) fact[i]=fact[i-1]*i%MOD; qword ans=C(n+m,n)-C(n+m,n+1); ans=(ans%MOD+MOD)%MOD; printf("%d ",(int)ans); }
Tarjan-缩点Tarjan-边双&&非递归Tarjan-点双整体二分
//bzoj 3339 #include<iostream> #include<cstring> #include<algorithm> #include<cstdio> #include<vector> #include<set> using namespace std; #define MAXN 210000 #define smid ((l+r)>>1) int n,m; int a[MAXN]; int tmp[MAXN],prv[MAXN]; const int big=200000; struct qur_t { int id,l,r,ans; }qur[MAXN]; vector<qur_t> vec; bool cmp_r(qur_t q1,qur_t q2) { return q1.r<q2.r; } int res[MAXN]; void solve(int l,int r,vector<qur_t> &vec) { if (l==r) { for (int i=0;i<vec.size();i++) res[vec[i].id]=l; return ; } vector<qur_t> v1,v2; multiset<int> S; sort(vec.begin(),vec.end(),cmp_r); for (int i=l;i<=smid;i++) S.insert(tmp[i]); int cur=n; for (int i=vec.size()-1;i>=0;i--) { while (vec[i].r<cur) { if (a[cur]<=smid && a[cur]>=l) { S.erase(S.find(cur)); S.insert(prv[cur]); } cur--; } if (*S.begin()<vec[i].l) v1.push_back(vec[i]); else v2.push_back(vec[i]); } solve(l,smid,v1); solve(smid+1,r,v2); } int main() { freopen("input.txt","r",stdin); freopen("output.txt","w",stdout); int x,y,z; scanf("%d%d",&n,&m); for (int i=1;i<=n;i++) scanf("%d",&a[i]); for (int i=1;i<=n;i++) { prv[i]=tmp[a[i]]; tmp[a[i]]=i; } for (int i=1;i<=m;i++) { scanf("%d%d",&qur[i].l,&qur[i].r); qur[i].id=i; } vec.insert(vec.begin(),qur+1,qur+m+1); solve(0,big+1,vec); for (int i=1;i<=m;i++) printf("%d ",res[i]); }
SBT && 动态凸包