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。

View Code

#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;

}