Sicily 2012. King 解题报告

题目：

Constraints

Time Limit: 1 secs, Memory Limit: 256 MB , Special Judge

Description

 There are n children in a country marked by integers from 1 to n.

They often fight with each other. For each pair of child A and child B, either A beats B or B beats A. It is not possible that both A beats B and B beats A. Of course, one never fight with himself.

Child A is not afraid of child B if A can beat B.

Child A is not afraid of child B if A can beat C and C can beat B either. Because A can say "I will call C to beat you" to B.

A child is called a king of children if he is not afraid of any other child.

Give you the beating relations.Find a king.

Input

 The first line contains a integer n which is between 1 and 1000. The following n lines contains n characters respectively. Character is either '0' or '1'. The Bth character of (A+1)th line will be '1' if and only if A can beat B. Input is terminated by EOF.

Output

 A number representing a king of children on a line. If such a king does not exist, output -1. If there are multiple kings, any one is accepted.

Sample Input

2
01
00

Sample Output

1

思路：

题目比较简单，从n个孩子中只要找出一个不怕任何一个孩子的即可。

对于第i个孩子，如果他与第j个孩子的关系为1说明他能打过j当然不怕他，如果关系为0则从第i个孩子能打过的人中找出一个能打过j的人。

#include<iostream>
using namespace std;

bool isTheKing(int i,string *beatRelations,int n);
bool isNotAfraidOf(string *beatRelations,int i,int j,int n);

int main(){
    int n;
    while(cin>>n){ //C++用循环输入可以实现题目的eof结束输入
        string beatRelations[n];
        for(int i=0;i<n;i++){
            cin>>beatRelations[i];
        }
        for(int i=0;i<n;i++){
            if(i==n-1){
                if(isTheKing(i,beatRelations,n)){
                    cout<<n<<endl;
                }else
                    cout<<-1<<endl; //最后一个孩子仍然不是king时要输出-1
            }else{
                if(isTheKing(i,beatRelations,n)){
                    cout<<i+1<<endl;
                    break;
                }
            }
        }
    }
    return 0;
}

bool isTheKing(int i,string *beatRelations,int n){
    for(int j=0;j<n;j++){ //遇到他害怕的孩子说明不是king即返回false
        if(beatRelations[i][j]=='0'&&i!=j){
            if(!isNotAfraidOf(beatRelations,i,j,n)) //beatRelations[i][j]=='0'要进行进一步的判断
                return false;
        }
    }
    return true;
}
bool isNotAfraidOf(string *beatRelations,int i,int j,int n){ //从他能打过的孩子中找能打过第j个孩子的人
    for(int k=0;k<n;k++){
        if(beatRelations[i][k]=='1'&&beatRelations[k][j]=='1')
            return true;
    }
    return false;
}