hdu1232 畅通工程 && poj2524 Ubiquitous Religions（并查集裸）

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http://acm.hdu.edu.cn/showproblem.php?pid=1232

题意：给你n个点m条边，求再需要添加多少条边就可以变成一个连通图。

思路：并查集裸体，思路看并查集详解。

#include <stdio.h>#include <algorithm>#include <stdlib.h>#include <string.h>#include <iostream>using namespace std;typedef long long LL;const int N = 100010;const int INF = 0x3f3f3f3f;int pre[N], sum;int Find(int x){    int r = x;    while(r != pre[r])        r = pre[r];    int i = x, j;    while(pre[i] != r)    {        j = pre[i];        pre[i] = r;        i = j;    }    return r;}void Union(int p1, int p2){    int x = Find(p1);    int y = Find(p2);    if(x != y)    {        pre[x] = y;        sum--;    }}int main(){   // freopen("in.txt", "r", stdin);    int n, m, p1, p2;    while(~scanf("%d", &n))    {        if(n == 0) break;        scanf("%d", &m);        sum = n-1;//原本的边数        for(int i = 1; i <= n; i++)            pre[i] = i;        for(int i = 1; i <= m; i++)        {            scanf("%d%d", &p1, &p2);            Union(p1, p2);        }        printf("%d\n", sum);    }    return 0;}

http://poj.org/problem?id=2524

题意：给你n个节点m个关系，每个关系代表两人在一个宗教，求宗教数量。

思路：刚开始连通分量，每加一条边减一个，最后即为所得。

#include <stdio.h>#include <algorithm>#include <stdlib.h>#include <string.h>#include <iostream>using namespace std;typedef long long LL;const int N = 100010;const int INF = 0x3f3f3f3f;int pre[N], sum;int Find(int x){    int r = x;    while(r != pre[r])        r = pre[r];    int i = x, j;    while(pre[i] != r)    {        j = pre[i];        pre[i] = r;        i = j;    }    return r;}void Union(int p1, int p2){    int x = Find(p1);    int y = Find(p2);    if(x != y)    {        pre[x] = y;        sum--;    }}int main(){   // freopen("in.txt", "r", stdin);    int n, m, p1, p2, Case = 1;    while(~scanf("%d%d", &n, &m))    {        if(n == 0 && m == 0) break;        sum = n;//原本的连通分量数        for(int i = 1; i <= n; i++)            pre[i] = i;        for(int i = 1; i <= m; i++)        {            scanf("%d%d", &p1, &p2);            Union(p1, p2);        }        printf("Case %d: %d\n", Case++, sum);    }    return 0;}

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