POJ1129Channel Allocation[迭代加深搜索　四色定理]

//
//  main.cpp
//  poj2157
//
//  Created by Candy on 9/30/16.
//  Copyright © 2016 Candy. All rights reserved.
//

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>
using namespace std;
const int N=28;
int n;
char s[N];
struct edge{
    int v,ne;
}e[N*N];
int h[N],cnt=0;
inline void ins(int u,int v){
    cnt++;
    e[cnt].v=v;e[cnt].ne=h[u];h[u]=cnt;
}
int col[N],num=0,has[N],maxcol,flag;
inline bool check(int u){
    for(int i=h[u];i;i=e[i].ne)
        if(col[e[i].v]==col[u]) return 0;
    return 1;
}
void dfs(int d){
    if(d>n) {flag=1;return;}
    if(flag) return;
    for(int cur=1;cur<=maxcol;cur++){
        col[d]=cur;
        if(check(d)) dfs(d+1);
        col[d]=0;
    }
    
}
int main(int argc, const char * argv[]) {
    while(scanf("%d",&n)!=EOF&&n){
        cnt=0;memset(h,0,sizeof(h));memset(col,0,sizeof(col));
        for(int i=1;i<=n;i++){
            scanf("%s",s+1);
            int len=strlen(s+1);int u=s[1]-‘A‘+1;
            for(int j=3;j<=len;j++){
                ins(u,s[j]-‘A‘+1);
            }
        }
        for(maxcol=1;maxcol<=4;maxcol++){
            flag=0;
            dfs(1);
            if(flag){
                if(maxcol==1)printf("1 channel needed.\n");
                else printf("%d channels needed.\n",maxcol);
                break;
            }
        }
    }
    
    
}