POJ 1129 四色问题 暴力搜索

#include <iostream>#include <algorithm>#include <string>#define N 30#define f(i,a,b) for(i=a;i<b;i++)//简化代码using namespace std;//图论中的染色问题，使得相邻两个节点颜色不同最少需要几种颜色//搜索就是暴力的枚举，记住bool maps[N][N];int n,i,j,k,q;char ch[N];void solve(){//如果出现四个区域，每个区域都和其他三个邻接,试探C42次，用四种颜色f(i,0,n)f(j,0,n)f(k,0,n)f(q,0,n)if(maps[i][j]&&maps[i][k] &&maps[i][q]&&maps[j][k] &&maps[j][q]&&maps[k][q]) { cout<<"4 channels needed."<<endl; return;}//如果有三个区域两两邻接，试探C32次，要用三种颜色 f(i,0,n)f(j,0,n)f(k,0,n) if(maps[i][j]&& maps[i][k] && maps[j][k]) { cout<<"3 channels needed."<<endl; return;}//如果存在两个区域邻接的情况，试探C22次，要用2种颜色 f(i,0,n)f(j,0,n)if(maps[i][j]) { cout << "2 channels needed." << endl; return;}//没有区域邻接 cout << "1 channel needed." << endl; }int main(){while(cin>>n,n!=0){memset(maps,0,sizeof(maps));f(i,0,n){cin>>ch;f(j,2,strlen(ch)){maps[ch[0]-'A'][ch[j]-'A'] = 1;maps[ch[j]-'A'][ch[0]-'A'] = 1;}}solve();}return 0;}