poj 1129 Channel Allocation（四色定理）

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<queue>
#include<algorithm>
#include<map>
#include<iomanip>
#include<climits>
#include<string.h>
#include<numeric>
#include<cmath>
#include<stdlib.h>
#include<vector>
#include<stack>
#include<set>
#define FOR(x, b, e)  for(int x=b;x<=(e);x++)
#define REP(x, n)     for(int x=0;x<(n);x++)
#define INF 1e7
#define MAXN 100010
#define maxn 1000010
#define Mod 1000007
#define N 33
using namespace std;
typedef long long LL;


int G[N][N];
int col[N];
int n, cnt;
bool d[N];

int main()
{
    string tmp;
    while (cin >> n, n) {
        memset(G, 0, sizeof(G));
        memset(col, 0, sizeof(col));
        for (int k = 0; k < n; ++k) {
            cin >> tmp;
            for (int i = 0; tmp[i + 2]; ++i)
                G[tmp[0] - 'A' + 1][tmp[i + 2] - 'A' + 1] = 1;
        }

        cnt = 0;
        int i, j;
        for (i = 1; i <= n; ++i) {
            memset(d,0,sizeof(d));
            for (j = 1; j <= i; ++j)        //寻找有冲突的频道编号 
                if (G[i][j]) d[col[j]] = 1;    //d[k]==1表示k频道有冲突 
            for (j = 1;; j++)                //寻找可以使用的编号最小的频道 
                if (d[j] == 0) break;
            col[i] = j;
            cnt = max(cnt, j);
        }
        if (cnt == 1)
            printf("%d channel needed.\n",cnt);
        else 
            printf("%d channels needed.\n", cnt);
    }
    return 0;
}

< / pre><pre name = "code" class = "cpp">#include "stdio.h"
#include "string.h"
#include "algorithm"
using namespace std;
int map[26][26];
int used[26];
char ch[30];
int n;
/* 这个dfs是基于四色定理，所以dfs第一层循环i最大为4 
   这种写法可以运作在四色定理的题目里，无向图任意的相邻两点颜色不同 

   */
void dfs(int node){
    int flag, i, j;
    if (node == n)
        return;
    for (i = 1; i <= 4; i++){
        flag = 0;
        //如果新结点与相邻结点颜色相同 则需要加颜色了，即i加1 
        for (j = 0; j<n; j++){
            if (map[node][j] && used[j] == i){
                flag = 1;
                break;
            }
        }
        //这个if意思是如果新结点与相邻结点颜色都不相同，那么就染旧颜色，再搜索下一个结点 
        if (!flag){
            used[node] = i;
            dfs(node + 1);
            break;
        }
    }
}

int main(){
    int i, j;
    while (scanf("%d", &n), n){
        memset(map, 0, sizeof(map));
        memset(used, 0, sizeof(used));
        for (i = 0; i < n; i++){
            scanf("%s", ch);
            for (j = 2; j < strlen(ch); j++)
            {
                map[i][ch[j] - 'A'] = 1;
                map[ch[j] - 'A'][i] = 1;
            }
        }
        dfs(0);
        sort(used, used + 26);
        if (used[25] == 1)
            printf("1 channel needed.\n");
        else
            printf("%d channels needed.\n", used[25]);
    }
}