• bzoj 2120: 数颜色


    暴力,,最暴力。。。。当然,用分块做也是可以的,用带修莫队也是可以的

     1 #include<bits/stdc++.h>
     2 #define N 100005
     3 #define LL long long
     4 #define inf 0x3f3f3f3f
     5 using namespace std;
     6 inline int ra()
     7 {
     8     int x=0,f=1; char ch=getchar();
     9     while (ch<'0' || ch>'9') {if (ch=='-') f=-1; ch=getchar();}
    10     while (ch>='0' && ch<='9') {x=x*10+ch-'0'; ch=getchar();}
    11     return x*f;
    12 }
    13 int n,m,T,f[11005],a[10005],cnt;
    14 map<int ,int > mp;
    15 int main()
    16 {
    17     n=ra(); m=ra();
    18     for (int i=1; i<=n; i++)
    19     {
    20         int x;
    21         if (!mp[x=ra()])
    22             mp[x]=++cnt;
    23         a[i]=mp[x];
    24     }
    25     for (int i=1; i<=m; i++)
    26     {
    27         char s[3]; scanf("%s",s);
    28         if (s[0]=='Q')
    29         {
    30             int ans=0; T++;
    31             int x=ra(),y=ra();
    32             for (int j=x; j<=y; j++)
    33                 if (f[a[j]]!=T) 
    34                     ans++,f[a[j]]=T;
    35             printf("%d
    ",ans);
    36         }
    37         else
    38         {
    39             int pos=ra(),col=ra();
    40             if (!mp[col])
    41                 mp[col]=++cnt;
    42             a[pos]=mp[col];
    43         }
    44     }
    45     return 0;
    46 }
     1 #include<bits/stdc++.h>
     2 #define N 1000005
     3 #define LL long long
     4 #define inf 0x3f3f3f3f
     5 using namespace std;
     6 inline int ra()
     7 {
     8     int x=0,f=1; char ch=getchar();
     9     while (ch<'0' || ch>'9') {if (ch=='-') f=-1; ch=getchar();}
    10     while (ch>='0' && ch<='9') {x=x*10+ch-'0'; ch=getchar();}
    11     return x*f;
    12 }
    13 struct node{int pos,col,pre;}w[N];
    14 struct data{int l,r,id,time;}q[N];
    15 int n,m,pos[N],last[N],col[N],cnt,sz,num[N],ans[N],ans1;
    16 bool vis[N];
    17 int cmp(data a, data b)
    18 {
    19     if (pos[a.l]==pos[b.l])
    20     {
    21         if (a.r==b.r) return a.time<b.time;
    22         return a.r<b.r;
    23     }
    24     return pos[a.l]<pos[b.l];
    25 }
    26 void cal(int x)
    27 {
    28     if (vis[x])
    29     {
    30         if (!--num[col[x]]) ans1--;
    31     }
    32     else 
    33     {
    34         if (++num[col[x]]==1) ans1++;
    35     }
    36     vis[x]^=1;
    37 }
    38 void change(int x, int c)
    39 {
    40     if (vis[x])
    41     {
    42         cal(x);
    43         col[x]=c;
    44         cal(x);
    45     }
    46     else col[x]=c;
    47 }
    48 int main()
    49 {
    50     n=ra(); m=ra();
    51     for (int i=1; i<=n; i++) col[i]=ra(),last[i]=col[i];
    52     for (int i=1; i<=m; i++)
    53     {
    54         char s[3]; scanf("%s",s);
    55         int x=ra(),y=ra();
    56         if (s[0]=='R')
    57         {
    58             w[++cnt].pos=x; w[cnt].col=y; w[cnt].pre=last[x];
    59             last[x]=y;
    60         }
    61         else
    62         {
    63             q[++sz].l=x; q[sz].r=y; q[sz].id=sz; q[sz].time=cnt;
    64         }
    65     }
    66     int block=sqrt(n);
    67     for (int i=1; i<=n; i++)
    68         pos[i]=(i-1)*block+1;
    69     sort(q+1,q+sz+1,cmp);
    70     int l=1,r=1; cal(1);
    71     for (int i=1; i<=m; i++)
    72     {
    73         for (int j=q[i-1].time+1; j<=q[i].time; j++) change(w[j].pos,w[j].col);
    74         for (int j=q[i-1].time; j>q[i].time; j--) change(w[j].pos,w[j].pre);
    75         while (l<q[i].l) cal(l++);
    76         while (l>q[i].l) cal(--l);
    77         while (r<q[i].r) cal(++r);
    78         while (r>q[i].r) cal(r--);
    79         ans[q[i].id]=ans1;
    80     }
    81     for (int i=1; i<=sz; i++) printf("%d
    ",ans[i]);
    82     return 0;
    83 }
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  • 原文地址:https://www.cnblogs.com/ccd2333/p/6482357.html
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