poj1129

题意：平面内最多26个点，给出一些点与点之间的矛盾关系，问最少使用多少颜色才能给这些点染色，并保证矛盾点之间不同色。

分析：深搜，对于每个点先尝试已经给别的点染过的颜色，如果不行再尝试染新的颜色，注意新的颜色只需要尝试一次即可，因为各种新的颜色对于当前状态来说是等价的。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
using namespace std;

#define maxn 30
#define maxm maxn * maxn
#define maxl maxn + 2
#define inf 0x3f3f3f3f

struct Edge
{
    int v, next;
}edge[maxm];

int head[maxn];
int color[maxn];
int n;
int edge_cnt;
int ans;

void addedge(int a, int b)
{
    edge[edge_cnt].v = b;
    edge[edge_cnt].next = head[a];
    head[a] = edge_cnt++;
}

void input()
{
    edge_cnt = 0;
    memset(head, -1, sizeof(head));
    char st[maxl];
    for (int i = 0; i < n; i++)
    {
        scanf("%s", st);
        int len = strlen(st);
        for (int j = 2; j < len; j++)
        {
            addedge(i, st[j] - 'A');
            addedge(st[j] - 'A', i);
        }
    }
}

bool ok(int a)
{
    for (int i = head[a]; ~i; i = edge[i].next)
        if (color[a] == color[edge[i].v])
            return false;
    return true;
}

void dfs(int color_num, int a)
{
    if (a == n)
    {
        ans = min(ans, color_num);
        return;
    }
    for (int i = 0; i < color_num; i++)
    {
        color[a] = i;
        if (ok(a))
            dfs(color_num, a + 1);
    }
    color[a] = color_num;
    dfs(color_num + 1, a + 1);
    color[a] = -1;
}

int main()
{
    //freopen("t.txt", "r", stdin);
    while (scanf("%d", &n), n)
    {
        input();
        ans = inf;
        memset(color, -1, sizeof(color));
        dfs(0, 0);
        if (ans == 1)
            printf("1 channel needed.\n");
        else
            printf("%d channels needed.\n", ans);
    }
    return 0;
}