HDU - 3342 floyd判环

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Resurgence_zc

最新推荐文章于 2024-08-10 21:23:20 发布

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分类专栏：最短路文章标签： acm 最短路

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本文介绍了一种通过Floyd算法来判断给定有向图中是否存在非法环的方法，并提供了具体的实现代码。

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HDU - 3342 Legal or Not

题目：

ACM-DIY is a large QQ group where many excellent acmers get together. It is so harmonious that just like a big family. Every day,many “holy cows” like HH, hh, AC, ZT, lcc, BF, Qinz and so on chat on-line to exchange their ideas. When someone has questions, many warm-hearted cows like Lost will come to help. Then the one being helped will call Lost “master”, and Lost will have a nice “prentice”. By and by, there are many pairs of “master and prentice”. But then problem occurs: there are too many masters and too many prentices, how can we know whether it is legal or not?

We all know a master can have many prentices and a prentice may have a lot of masters too, it’s legal. Nevertheless，some cows are not so honest, they hold illegal relationship. Take HH and 3xian for instant, HH is 3xian’s master and, at the same time, 3xian is HH’s master,which is quite illegal! To avoid this,please help us to judge whether their relationship is legal or not.

Please note that the “master and prentice” relation is transitive. It means that if A is B’s master ans B is C’s master, then A is C’s master.
Input
The input consists of several test cases. For each case, the first line contains two integers, N (members to be tested) and M (relationships to be tested)(2 <= N, M <= 100). Then M lines follow, each contains a pair of (x, y) which means x is y’s master and y is x’s prentice. The input is terminated by N = 0.
TO MAKE IT SIMPLE, we give every one a number (0, 1, 2,…, N-1). We use their numbers instead of their names.
Output
For each test case, print in one line the judgement of the messy relationship.
If it is legal, output “YES”, otherwise “NO”.

题意：

给一个有向图，让你去判断是否存在环；

思路：

所谓有环就是存在两个点a1，a2，a1->a2和a2->a1,都可通，
那么就是进行一次floyd算法，每次更新dp[i][j],的时候都去判断一下dp[j][i]是否不为inf，如果有就是存在环，否则不存在环。
wa点的话，强说一个是i==j时当然要跳过，一个是maps[i][i]=0;(这个是因为，就算不是0一样可以得到最短路,但是会有情况判不到)。

代码：

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define inf 0x3f3f3f3f
using namespace std;
int n,m;
int maps[110][110];
bool floyd()
{
    for(int k=0;k<n;k++)
    {
        for(int i=0;i<n;i++)
        {
            if(maps[i][k]==inf)
                continue;
            for(int j=0;j<n;j++)
            {
                if(i==j||maps[k][j]==inf)
                    continue;
                if(maps[i][j]>maps[i][k]+maps[k][j])
                    maps[i][j]=maps[i][k]+maps[k][j];
                if(maps[j][i]!=inf)
                    return false;
            }
        }
    }
    return true;
}
int main()
{
    while(scanf("%d%d",&n,&m),n)
    {
        memset(maps,inf,sizeof(maps));
        int x,y;
        for(int i=1;i<=m;i++)
        {
            scanf("%d%d",&x,&y);
            maps[x][y]=1;
        }
        for(int i=0;i<n;i++)
            maps[i][i]=0;
        if(floyd())
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}

P.S.迷之切水题日常orz