四色定理+dfs（poj 1129）

　　题目：Channel Allocation

　　题意：要求A:BCD，A与B,C,D都不相同，求不同的值，典型的四色定理；

　　

#include <iostream>
#include <algorithm>
#include <stdlib.h>
#include <time.h>
#include <cmath>
#include <cstdio>
#include <string>
#include <cstring>
#include <vector>
#include <queue>
#include <stack>
#include <set>

#define c_false ios_base::sync_with_stdio(false); cin.tie(0)
#define INF 0x3f3f3f3f
#define INFL 0x3f3f3f3f3f3f3f3f
#define zero_(x,y) memset(x , y , sizeof(x))
#define zero(x) memset(x , 0 , sizeof(x))
#define MAX(x) memset(x , 0x3f ,sizeof(x))
#define swa(x,y) {LL s;s=x;x=y;y=s;}
using namespace std ;
#define N 105

const double PI = acos(-1.0);
typedef long long LL ;

int mapp[N][N], color[N];
int col,flag;
int n;
string s;
bool ok(int i){
    for(int j = ; j <= ; j++){
        if(!mapp[i][j]) continue;
        if(color[i] == color[j]) return false;
    }
    return true;
}

void dfs(int num){
    if(num > n) {flag = ; return ;}
    for(int i = ; i <= col; i ++){
        color[num] = i;
        if(ok(num))
            dfs(num+);
        color[num]= ;    ///记得回溯时要清零啊！！！
    }
}

int main(){
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    while(cin>>n && n){
        flag = ;
        zero(mapp);
        for(int i = ; i< n;i++){
            cin>>s;
            int k = s.size();
            int x,y;
            for(int j = ;j< k;j++){
                x = s[] - 'A' +;
                y = s[j] - 'A' +;
                mapp[x][y] = mapp[y][x] = ;
            }
        }
        for(col = ; col <= ; col++){
            dfs();
            if(flag) break;
        }
        if(col == )
            printf("%d channel needed.\n", col);
        else
            printf("%d channels needed.\n", col);
    }
    return ;
}