P2850 [USACO06DEC]虫洞Wormholes

题目描述

While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it is a one-way path that delivers you to its destination at a time that is BEFORE you entered the wormhole! Each of FJ's farms comprises N (1 ≤ N ≤ 500) fields conveniently numbered 1..N, M (1 ≤ M ≤ 2500) paths, and W (1 ≤ W ≤ 200) wormholes.

As FJ is an avid time-traveling fan, he wants to do the following: start at some field, travel through some paths and wormholes, and return to the starting field a time before his initial departure. Perhaps he will be able to meet himself :) .

To help FJ find out whether this is possible or not, he will supply you with complete maps to F (1 ≤ F ≤ 5) of his farms. No paths will take longer than 10,000 seconds to travel and no wormhole can bring FJ back in time by more than 10,000 seconds.

John在他的农场中闲逛时发现了许多虫洞。虫洞可以看作一条十分奇特的有向边，并可以使你返回到过去的一个时刻（相对你进入虫洞之前）。John的每个农场有M条小路（无向边)连接着N （从1..N标号）块地，并有W个虫洞。其中1<=N<=500,1<=M<=2500,1<=W<=200。现在John想借助这些虫洞来回到过去（出发时刻之前），请你告诉他能办到吗。 John将向你提供F(1<=F<=5)个农场的地图。没有小路会耗费你超过10000秒的时间，当然也没有虫洞回帮你回到超过10000秒以前。

输入格式

Line 1: A single integer, F. F farm descriptions follow.

Line 1 of each farm: Three space-separated integers respectively: N, M, and W

Lines 2..M+1 of each farm: Three space-separated numbers (S, E, T) that describe, respectively: a bidirectional path between S and E that requires T seconds to traverse. Two fields might be connected by more than one path.

Lines M+2..M+W+1 of each farm: Three space-separated numbers (S, E, T) that describe, respectively: A one way path from S to E that also moves the traveler back T seconds.

输出格式

Lines 1..F: For each farm, output "YES" if FJ can achieve his goal, otherwise output "NO" (do not include the quotes).

输入输出样例

输入 #1

2
3 3 1
1 2 2
1 3 4
2 3 1
3 1 3
3 2 1
1 2 3
2 3 4
3 1 8

输出 #1

NO
YES

说明/提示

For farm 1, FJ cannot travel back in time.

For farm 2, FJ could travel back in time by the cycle 1->2->3->1, arriving back at his starting location 1 second before he leaves. He could start from anywhere on the cycle to accomplish this.

【思路】

SFPA判负环
很水的一道SPFA判负环的板子题

【题目分析】

在农田没有虫洞的道路上面走回消耗时间
在有虫洞的道路上面会回到之前的时间
也就是得到时间
所以会使之前消耗的时间减少
只要消耗的时间能够成为负数
就是回到了出发时间之间
能够成为负数
也就是权值为负的一个环
就是负环
所以只需要判断有没有负环就可以了

【SPFA判负环】

板子
赤裸裸的一道板子题
详情解释见
这里

【完整代码】

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
using namespace std;
const int Max = 6004;
const int M = 505;
struct node
{
    int y,ne,z;
}a[Max];
int n,m,w; 
int head[M],sum = 0;

void add(int x,int y,int z)
{
    a[++ sum].y = y;
    a[sum].ne = head[x];
    a[sum].z = z;
    head[x] = sum;
}

int d[M],cnt[M];
bool use[M];
bool SPFA()
{
    memset(cnt,0,sizeof(cnt));
    memset(use,false,sizeof(use));
    for(register int i = 1;i <= n;++ i)
        d[i] = 999999;
    d[1] = 0;
    queue<int>q;
    q.push(1);
    while(!q.empty())
    {
        int qwq = q.front();
        q.pop();use[qwq] = false;
        for(register int i = head[qwq];i != 0;i = a[i].ne)
        {
            int awa = a[i].y;
            if(d[awa] > d[qwq] + a[i].z)
            {
                d[awa] = d[qwq] + a[i].z;
                cnt[awa] = cnt[qwq] + 1;
                if(cnt[awa] > n)
                    return false;
                if(use[awa] == false)
                {
                    use[awa] = true;
                    q.push(awa);
                }
            }
        }
    }
    return true;
}

int main()
{
    int f;
    scanf("%d",&f);
    while(f --)
    {
        sum = 0;
        memset(head,0,sizeof(head));
        scanf("%d%d%d",&n,&m,&w);
        int x,y,z;
        for(register int i = 1;i <= m;++ i)
        {
            scanf("%d%d%d",&x,&y,&z);
            add(x,y,z);add(y,x,z);
        }
        for(register int i = 1;i <= w;++ i)
        {
            scanf("%d%d%d",&x,&y,&z);
            add(x,y,-z);
        }
        if(SPFA() == false)
            cout << "YES" << endl;
        else
            cout << "NO" << endl;
    }
    return 0;
}

洛谷 P2850 [USACO06DEC]虫洞Wormholes 题解