D-City

5 10 
0 1 
1 2 
1 3 
1 4 
0 2 
2 3 
0 4 
0 3 
3 4 
2 4

1 
1 
1 
2 
2 
2 
2 
3 
4 
5
Hint
The graph given in sample input is a complete graph, that each pair of vertex has an edge connecting them, so there‘s only 1 connected block at first. 
The first 3 lines of output are 1s  because  after  deleting  the  first  3  edges  of  the  graph,  all  vertexes  still  connected together. 
But after deleting the first 4 edges of the graph, vertex 1 will be disconnected with other vertex, and it became an independent connected block. 
Continue deleting edges the disconnected blocks increased and finally it will became the number of vertex, so the last output should always be N. 
 

题意：给你一个n个点的图，任意两点之间都有一条无向边，然后Q个查询

          每个查询求删除一条边后剩下多少块。

题解：边删完了块数肯定是n,所以我们可以从后面开始，一次加入边，然后求块数。

#include<cstring>
#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cmath>
#include<vector>
#define N 10010
#define M 100100
#define ll long long

using namespace std;

int par[N];
int r[N];
int s[M];
int n,m;

struct node {
    int x,y;
} a[M];

void init() {
    for(int i=0; i<=n; i++) {
        par[i]=i;
        r[i]=0;
    }
}

int finds(int x) {
    if(par[x]==x)return x;
    return par[x]=finds(par[x]);
}

void unite(int x,int y) {
    x=finds(x);
    y=finds(y);
    if(x==y)return;
    if(r[x]<r[y]) {
        par[x]=y;
    } else {
        par[y]=x;
        if(r[x]==r[y])r[x]++;
    }
}

bool same(int x,int y) {
    return finds(x)==finds(y);
}

int main() {
    //freopen("test.in","r",stdin);
    while(cin>>n>>m) {
        for(int i=0; i<m; i++) {
            scanf("%d%d",&a[i].x,&a[i].y);
        }
        init();
        int ans=n;
        for(int i=m-1; i>=0; i--) {
            s[i]=ans;
            int x=a[i].x,y=a[i].y;
            if(!same(x,y)) {
                ans--;
            }
            unite(x,y);
        }
        for(int i=0; i<m; i++) {
            printf("%d\n",s[i]);
        }
    }
    return 0;
}

hdu 4496 D-City(并查集)

原文：http://blog.csdn.net/acm_baihuzi/article/details/46861325