2015 多校联赛 ——HDU5305（搜索）

Friends

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

 Total Submission(s): 163    Accepted Submission(s): 61

Problem Description

There are n people and m pairs of friends. For every pair of friends, they can choose to become online friends (communicating using online applications) or offline friends (mostly using face-to-face communication). However, everyone in these n people wants to have the same number of online and offline friends (i.e. If one person has x onine friends, he or she must have x offline friends too, but different people can have different number of online or offline friends). Please determine how many ways there are to satisfy their requirements. 

 

Sample Input

23 31 22 33 14 41 22 33 44 1

 

Sample Output

02

 

Source

2015 Multi-University Training Contest 2

求：朋友关系可离线可在线，要求每个人的离线朋友数等于在线朋友数，总共有多少种方法

先对入度判断，奇数直接不可能，然后再进行搜索

#include <iostream>#include <cstring>#include <cstdio>#include <queue>#include <vector>using namespace std;struct node{    int u,v;} pnode[50];int in[10];int c[10],c2[10];int ans,n,m;int num,tem;void dfs(int u){    if(u == m+1)    {        for(int i = 1; i <= n; i++)            if(c[i] != c2[i])                return;        ans++;        return ;    }    int a = pnode[u].u;    int b = pnode[u].v;    if(c[a] < in[a]/2 && c[b] < in[b]/2)    {        c[a] ++;        c[b] ++;        dfs(u+1);        c[a]--;        c[b]--;    }    if(c2[a] < in[a]/2 && c2[b] < in[b]/2)    {        c2[a] ++;        c2[b] ++;        dfs(u+1);        c2[a]--;        c2[b]--;    }    return;}int main(){    int T;    scanf("%d",&T);    while(T--)    {        int flag = 0;        scanf("%d%d",&n,&m);        memset(in,0,sizeof(in));        for(int i = 1; i <= m; i++)        {            scanf("%d%d",&pnode[i].u,&pnode[i].v);            in[pnode[i].u]++;            in[pnode[i].v]++;        }        memset(c,0,sizeof(c));        memset(c2,0,sizeof(c2));        for(int i = 1; i <= n; i++)        {            if(in[i] % 2 == 1)            {                flag = 1;                break;            }        }        if(flag){            printf("0\n");            continue;        }        ans = 0;        dfs(1);        printf("%d\n",ans);    }    return 0;}