[kuangbin带你飞]专题四最短路练习 F F - Wormholes

题目链接：https://vjudge.net/contest/66569#problem/F

题目：

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.

Input

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.

Output

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

Sample Input

Sample Output

NO
YES

Hint

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.

题意：

农夫约翰在探索他的许多农场，发现了一些惊人的虫洞。虫洞是很奇特的，因为它是一个单向通道，可让你进入虫洞的前达到目的地！他的N（1≤N≤500）个农场被编号为1..N，之间有M（1≤M≤2500）条路径，W（1≤W≤200）个虫洞。FJ作为一个狂热的时间旅行的爱好者，他要做到以下几点：开始在一个区域，通过一些路径和虫洞旅行，他要回到最开时出发的那个区域出发前的时间。也许他就能遇到自己了:)。为了帮助FJ找出这是否是可以或不可以，他会为你提供F个农场的完整的映射到（1≤F≤5）。所有的路径所花时间都不大于10000秒，所有的虫洞都不大于万秒的时间回溯。

Input

第1行：一个整数F表示接下来会有F个农场说明。每个农场第一行：分别是三个空格隔开的整数：N，M和W 第2行到M+1行：三个空格分开的数字（S，E，T）描述，分别为：需要T秒走过S和E之间的双向路径。两个区域可能由一个以上的路径来连接。第M +2到M+ W+1行：三个空格分开的数字（S，E，T）描述虫洞，描述单向路径，S到E且回溯T秒。

Output

F行，每行代表一个农场每个农场单独的一行，” YES”表示能满足要求，”NO”表示不能满足要求。

思路：

套判断负环的模板即可

先给出我一开始用的spfa算法如下，用时235s：

//
// Created by hanyu on 2019/7/19.
//
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
const int maxn=2e5+7;
#define MAX 0x3f3f3f3f
int book[maxn],cnt[maxn],d[maxn],head[maxn];
int n,m,w;
int pos;
struct Node{
    int s;
    int e;
    int t;
}node[maxn];
void add(int s,int e,int t)
{
    node[pos].s=e;
    node[pos].e=t;
    node[pos].t=head[s];
    head[s]=pos++;
}
bool spfa(int start)
{

    queue<int>qu;
    qu.push(start);
    book[start]=1;
    d[start]=0;
    while(!qu.empty())
    {
        int now=qu.front();
        qu.pop();
        book[now]=0;
        for(int i=head[now];i!=-1;i=node[i].t)
        {
            int ss=node[i].s;
            int ee=node[i].e;
            if(d[ss]>d[now]+ee)
            {
                d[ss]=d[now]+ee;
                if(!book[ss])
                {
                    qu.push(ss);
                    book[ss]=1;
                    cnt[ss]++;
                    if(cnt[ss]>=n)
                    return true;//判断负环
                }
            }
        }
    }
    return false;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        memset(book,0,sizeof(book));
        memset(cnt,0,sizeof(cnt));
        memset(d,MAX,sizeof(d));
        memset(head,-1,sizeof(head));
        memset(node,0,sizeof(node));
        scanf("%d%d%d",&n,&m,&w);
        int s,e,t;
        for(int i=1;i<=m;i++)
        {
            scanf("%d%d%d",&s,&e,&t);
            add(s,e,t);
            add(e,s,t);
        }
        for(int i=1;i<=w;i++)
        {
            scanf("%d%d%d",&s,&e,&t);
            add(s,e,-t);
        }
        cnt[1]=1;
        pos=0;
        if(spfa(1))
            printf("YES
");
        else
            printf("NO
");
    }
    return 0;
}

然后学习了优化版的Bellmanford算法，发现速度更快，用时79s

//
// Created by hanyu on 2019/7/20.
//
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
const int maxn=6005;
#define MAX 0x3f3f3f3f
int n,m,w;

int pos;
int d[maxn];

struct Node{
    int u,v,w;
}node[maxn];
void add(int u,int v,int w)
{
    node[++pos].u=u;
    node[pos].v=v;
    node[pos].w=w;
}
bool bell(){

    memset(d,MAX,sizeof(d));
    d[1]=0;
    int flag=0;
    for(int i=1;i<=n;i++)
    {
       flag=0;
       for(int j=1;j<=pos;j++)
       {
           if(d[ node[j].v]>d[node[j].u]+node[j].w)
           {
               d[ node[j].v]=d[node[j].u]+node[j].w;
               flag=1;
           }
       }
       if(flag==0)
           return false;
    }
    flag=0;
    for(int i=1;i<=pos;i++)
    {
        if(d[node[i].v]>d[node[i].u]+node[i].w)
            return 1;
    }
    return 0;
}
int main()
{
    int T;
    int a,b,c;
    scanf("%d",&T);
    while(T--)
    {
        pos=0;
        scanf("%d%d%d
",&n,&m,&w);
        for(int i=1;i<=m;i++)
        {
            scanf("%d%d%d",&a,&b,&c);
            add(a,b,c);
            add(b,a,c);
        }
        for(int i=1;i<=w;i++)
        {
            scanf("%d%d%d",&a,&b,&c);
            add(a,b,-c);
        }
        if(bell())
            printf("YES
");
        else
            printf("NO
");
    }
    return 0;
}

[kuangbin带你飞]专题四 最短路练习 F F - Wormholes

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[kuangbin带你飞]专题四最短路练习 F F - Wormholes