POJ 1129 Channel Allocation 四色定理dfs

题目： http://poj.org/problem?id=1129

开始没读懂题，看discuss的做法，都是循环枚举的，很麻烦。然后我就决定dfs，调试了半天终于0ms A了。

#include <stdio.h>
#include <string.h>
bool graph[26][26], vis[26][4];
int n, ans;

void calc()
{
    int cnt = 0;
    for(int i = 0; i < 4; i++)
    {
        for(int j = 0; j < n; j++)
        {
            if(vis[j][i])
            {
                cnt++;
                break;
            }
        }
    }
    if(cnt < ans)ans = cnt;
}

void dfs(int x)
{
    if(x >= n)
    {
        calc();
        return;
    }

    for(int i = 0; i < 4; i++)
    {
        bool ok = 1;
        for(int j = 0; j < n; j++)
        {
            if((graph[j][x] && vis[j][i]) || (graph[x][j] && vis[j][i]))
            {
                ok = 0;
                break;
            }
        }
        if(ok)
        {
            vis[x][i] = 1;
            dfs(x+1);
            vis[x][i] = 0;
        }
    }
}

int main()
{
    char s[30];
    while(scanf("%d", &n) != EOF && n)
    {
        ans = 0x3f3f3f3f;
        memset(graph, 0, sizeof(graph));
        memset(vis, 0, sizeof(vis));
        for(int i = 0; i < n; i++)
        {
            scanf("%s", s);
            for(int j = 2; s[j]; j++)
            {
                graph[i][s[j]-'A'] = 1;
            }
        }
        vis[0][0] = 1;
        dfs(1);
        if(ans == 1)
            printf("1 channel needed.
");
        else printf("%d channels needed.
", ans);
    }
    return 0;
}