poj 1129 Channel Allocation

可以转化为着色模型

dfs + 四色定理

#include<cstdio>
#include<memory.h>
int n,num;
int d[100][100];
int c[100];

bool ok(int step)
{
    for(int i = 0; i < n; i++)
        if(d[step][i] == 1 && c[step] == c[i]) return false;
    return true;
}

bool dfs(int num,int step)
{
    if(step >= n) return true;
    for(int i = 1; i <= num; i++)
    {
        c[step] = i;
        if(ok(step) && dfs(num,step+1)) return true;

        c[step] = 0;//恢复
    }
    return false;
}

int main()
{
    freopen("input.txt","r",stdin);
    while(scanf("%d
",&n) && n)
    {
        char ch;
        memset(d,0,sizeof(d));
        for(int i = 0; i < n; i++)
        {
            scanf("%c:",&ch);
            while(scanf("%c",&ch) && ch != '
')
                d[i][ch - 'A'] = d[ch - 'A'][i] = 1;
        }

        memset(c,0,sizeof(c));
        int num = 0;
        for(num = 1; num < 4; num++)
            if(dfs(num,0)) break;

        if(num == 1) printf("1 channel needed.
");
        else printf("%d channels needed.
",num);

    }
    return 0;
}