ZOJ 4124 2019 ACM山东省赛 L题

Median

Time Limit: 1 Second Memory Limit: 65536 KB

Recall the definition of the median of $elements where is odd: sort these elements and the median is the -th largest element.$

In this problem, the exact value of each element is not given, but $relations between some pair of elements are given. The -th relation can be described as, which indicates that the -th element is strictly larger than the -th element.$

For all $, is it possible to assign values to each element so that all the relations are satisfied and the -th element is the median of the elements?$

Input

There are multiple test cases. The first line of the input contains an integer $, indicating the number of test cases. For each test case:$

The first line contains two integers $and (,), indicating the number of elements and the number of relations. It's guaranteed that is odd.$

For the following $lines, the -th line contains two integers and, indicating that the -th element is strictly larger than the -th element. It guaranteed that for all, or .$

It's guaranteed that the sum of $of all test cases will not exceed .$

Output

For each test case output one line containing one string of length $. If it is possible to assign values to each element so that all the relations are satisfied and the -th element is the median, the -th character of the string should be '1', otherwise it should be '0'.$

Sample Input

Sample Output

01000

000

Hint

For the first sample test case, as the 2nd element is smaller than the 1st and the 3rd elements and is larger than the 4th and the 5th elements, it's possible that the 2nd element is the median.

For the second sample test case, as the 1st element can't be larger than itself, it's impossible to assign values to the elements so that all the relations are satisfied.

Author: WANG, Yucheng
Source: The 10th Shandong Provincial Collegiate Programming Contest

省赛时做完5道题之后就没有在做出来题了，真难受。

当时想用拓扑来找到比他大的有多少个和比他小的有多少个。可惜拓扑没写出来，真菜啊。

问了一下人家山理大一的新生，人家说用Floyd跑一下就出来了，emmm。。。。太菜了。菜哭了。

#include<cstdio>
#include<cstring>
#include<vector>
using namespace std;

int n;
int graph[105][105];
int v[2][105];
void print(){
    for(int i=0;i<n;i++)
        putchar('0');
    puts("");
}
int floyd(){
    for(int k=1;k<=n;k++) // floyd
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
                if(graph[i][k] && graph[k][j])
                    graph[i][j] = 1;

    for(int i=1;i<=n;i++){ // 判环
        for(int j=1;j<=n;j++){
            if(graph[i][j] && graph[j][i])
                return -1;
        }
    }
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++)if(graph[i][j]){ // graph[i][j] 表示 i严格大于j
            v[0][i]++; // v[0][i] 记录比 i 小的有多少
            v[1][j]++; // v[0][j] 记录比 j 大的有多少
        }
    }
    return 0;
}
int main(){
    int t, m;
    int x, y;
    scanf("%d",&t);
    while(t--){
        memset(graph,0,sizeof(graph));
        memset(v,0,sizeof(v));
        scanf("%d%d",&n,&m);
        for(int i=0;i<m;i++){
            scanf("%d%d",&x,&y);
            graph[x][y] = 1;
        }
        if(floyd()==-1){// 有环
            print();
            continue;
        }
        for(int i=1;i<=n;i++){
            if(v[0][i]<=n/2 && v[1][i]<=n/2)
                putchar('1');
            else
                putchar('0');
        }
        puts("");
    }
    return 0;
}

View Code