hdu 5305 Friends 【暴搜】

来源：互联网发布：端口的作用是什么编辑：程序博客网时间：2024/05/17 18:23

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5305

题意：给一个无向图，每条边可以是online边也可以是offline边，问
有多少种方法使得每个节点的online边和offline边一样多

解法：暴搜。记录每个点连接的边数，奇数的直接不可能，偶数的分成两个数组，c1[i]表示i的在线朋友数，c2[i]表示i的离线朋友数，然后一条边一条边搜就行了。

代码：

#include <stdio.h>#include <ctime>#include <math.h>#include <limits.h>#include <complex>#include <string>#include <functional>#include <iterator>#include <algorithm>#include <vector>#include <stack>#include <queue>#include <set>#include <map>#include <list>#include <bitset>#include <sstream>#include <iomanip>#include <fstream>#include <iostream>#include <ctime>#include <cmath>#include <cstring>#include <cstdio>#include <time.h>#include <ctype.h>#include <string.h>#include <assert.h>using namespace std;struct{    int u;    int v;}e[100];int c1[100], c2[100];int num[100];int ans;int n, m;void dfs(int x){    if (x == m + 1)    {        ans++;        return;    }    int u = e[x].u;    int v = e[x].v;    if (c1[u] && c1[v])    {        c1[u] --;        c1[v] --;        dfs(x + 1);        c1[u] ++;        c1[v] ++;    }    if (c2[u] && c2[v])    {        c2[u] --;        c2[v] --;        dfs(x + 1);        c2[u] ++;        c2[v] ++;    }    return;}int main(){    int t;    scanf("%d",&t);    while (t--)    {        memset(c1, 0, sizeof(c1));        memset(c2, 0, sizeof(c2));        memset(num, 0, sizeof(num));        scanf("%d%d",&n,&m);        for (int i = 1; i <= m;i++)        {            scanf("%d%d",&e[i].u,&e[i].v);            num[e[i].u]++;            num[e[i].v]++;        }        int ok = 1;        for (int i = 1; i <= n; i++)        {            c1[i] = c2[i] = num[i] / 2;            if (num[i] & 1)            {                ok = 0;                break;            }        }        if (!ok)        {            printf("0\n");            continue;        }        ans = 0;        dfs(1);        printf("%d\n",ans);    }    return 0;}

参考自：http://blog.csdn.net/Tc_To_Top/article/details/47027079

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