[CF1303F] Number of Components - disjoint-set time back

A \ (n \ times m \) the matrix, all initial state \ (0 \) .

If two adjacent elements (four communication) are equal, we say that they are in communication, and the relationship may be transferred.

There \ (Q \) operations, each specifying a location \ ((x_i, y_i) \ ) replace it with \ (C_i \) .

How many blocks each communication request after the matrix operation there.

\(q \leq 2\times 10^6\), \(n,m \leq 300\)

Solution

With deleted disjoint-set problem can be off-line, so positive with each doing a backwards, then the answer will be able to make the difference

Consider the process of being made, the beginning is a piece of \ (0 \) of the plate board

Every time we create a new node, and then attempt to merge it with the surrounding nodes, the number of merger set to \ (t \) , then this time contribution to the answers (that is, the operation of how many new communication block) is \ (1-t \)

Similarly, the contribution with a negative sign reverse operation on it

Finally, when the output of the answer, the contribution of the array and to seek prefix

#include <bits/stdc++.h>
using namespace std;

const int N = 305, M = 2000005;
int a[N][N],n,m,q,ind,num,f[M*2],ans[M],id[N][N];

struct query {
    int x,y,b,c;
} s[M];

int find(int x) {return (x==f[x])?x:f[x]=find(f[x]);}
void merge(int x,int y) {if((x=find(x))-(y=find(y))) f[x]=y,--num;}

void solve(int x,int y) {
    if(a[x][y]==a[x-1][y]) merge(id[x][y],id[x-1][y]);
    if(a[x][y]==a[x+1][y]) merge(id[x][y],id[x+1][y]);
    if(a[x][y]==a[x][y-1]) merge(id[x][y],id[x][y-1]);
    if(a[x][y]==a[x][y+1]) merge(id[x][y],id[x][y+1]);
}

signed main() {
    scanf("%d%d%d",&n,&m,&q);
    for(int i=1;i<=q;i++) {
        int x,y,c;
        scanf("%d%d%d",&x,&y,&c);
        s[i]={x,y,a[x][y],c};
        a[x][y]=c;
    }
    memset(a,0xff,sizeof a);
    for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) a[i][j]=0;
    for(int i=1;i<=q;i++) {
        if(s[i].b!=s[i].c) {
            int x=s[i].x,y=s[i].y,b=s[i].b,c=s[i].c;
            num=1;
            a[x][y]=c;
            id[x][y]=++ind;
            f[ind]=ind;
            solve(x,y);
            ans[i]+=num;
        }
    }
    ind=0;
    memset(f,0,sizeof f);
    for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) {
        id[i][j]=++ind;
        f[ind]=ind;
    }
    for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) solve(i,j);
    for(int i=q;i>=1;--i) {
        if(s[i].b!=s[i].c) {
            int x=s[i].x,y=s[i].y,b=s[i].c,c=s[i].b;
            num=1;
            a[x][y]=c;
            id[x][y]=++ind;
            f[ind]=ind;
            solve(x,y);
            ans[i]-=num;
        }
    }
    ans[0]=1;
    for(int i=1;i<=q;i++) ans[i]+=ans[i-1];
    for(int i=1;i<=q;i++) printf("%d\n",ans[i]);
}