Wormholes POJ - 3259（最短路，求是否存在负环）

POJ—3259

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
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
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.

t译文：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。作为一个狂热的时间旅行FJ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。

输入
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。
输出
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。
NO
YES
题意是问是否能通过虫洞回到过去；
虫洞是一条单向路，不但会把你传送到目的地，而且时间会倒退Ts。
我们把虫洞看成是一条负权路，问题就转化成求一个图中是否存在负权回路

下面给出三种算法的代码
floyd

#include
#include
#include
#include
using namespace std;
int map[505][505],n,m,k,num=0;
int floyd()
{
    int i,j,k,f=0;
    for(k=1;k<=n;k++)
        for(i=1;i<=n;i++){
            for(j=1;j<=n;j++)
            {
                int t=map[i][k]+map[k][j];
                if(map[i][j]>t)map[i][j]=t;
                /*map[i][j]=min(map[i][j],map[i][k]+map[k][j]);*/   //注意这里，用错了会超时！！！！不用min就跑了1688ms，，，，
            }
            if(map[i][i]<0)return 1;
    }
    return f;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int i,j,a,b,c;
        scanf("%d%d%d",&n,&m,&k);
        memset(map,0x3f3f3f3f,sizeof(map));
        for(i=1;i<=n;i++)map[i][i]=0;
        for(i=1;i<=m;i++)
        {
            scanf("%d%d%d",&a,&b,&c);
            if(c<map[a][b])map[a][b]=map[b][a]=c;
        }
        for(i=1;i<=k;i++)
        {
            scanf("%d%d%d",&a,&b,&c);
            map[a][b]=-c;
        }
        num++;
        int f=floyd();
        if(!f)printf("NO\n");
        else printf("YES\n");
    }
    return 0;
}

BellmanFord:

#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
#define N 5210
#define INF 0xfffffff

int cnt, dist[N], Head[N];
int n, m, w;

struct Edge
{
    int u, v, w, next;
}e[N];

void Add(int u, int v, int w)
{
    e[cnt].u = u;
    e[cnt].v = v;
    e[cnt].w = w;
    e[cnt].next = Head[u];
    Head[u] = cnt++;
}

bool BellmanFord()
{
    dist[1] = 0;
    for(int i=0; i<n; i++)
    {
        for(int j=0; j<cnt; j++)
        {
            if(dist[e[j].v] > dist[e[j].u]+e[j].w)
                dist[e[j].v] = dist[e[j].u]+e[j].w;
        }
    }
    for(int i=0; i<cnt; i++)
    {
        if(dist[e[i].v] > dist[e[i].u]+e[i].w)
            return 0;
    }
    return 1;
}

int main()
{
    int T, a, b, c;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d%d%d", &n, &m, &w);

        cnt = 0;
        memset(Head, -1, sizeof(Head));
        for(int i=1; i<=n; i++)
            dist[i] = INF;

        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( !BellmanFord() )
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}

spfa:

#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
#define N 5210
#define INF 0xfffffff

int cnt, dist[N], Head[N], num[N], vis[N];
int n, m, w;

struct Edge
{
    int v, w, next;
}e[N];

void Add(int u, int v, int w)
{
    e[cnt].v = v;
    e[cnt].w = w;
    e[cnt].next = Head[u];
    Head[u] = cnt++;
}

bool spfa()///spfa模板；
{
    memset(vis, 0, sizeof(vis));
    memset(num, 0, sizeof(num));
    queue<int>Q;
    vis[1] = 1;
    dist[1] = 0;
    Q.push(1);
    num[1]++;
    while(Q.size())
    {
        int p=Q.front();
        Q.pop();
        vis[p] = 0;
        for(int i=Head[p]; i!=-1; i=e[i].next)
        {
            int q = e[i].v;
            if(dist[q] > dist[p] + e[i].w)
            {
                dist[q] = dist[p] + e[i].w;
                if(!vis[q])
                {
                    vis[q] = 1;
                    Q.push(q);
                    num[q] ++;
                    if(num[q]>n)
                        return true;
                }
            }
        }
    }
    return false;
}

int main()
{
    int T, a, b, c;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d%d%d", &n, &m, &w);

        cnt = 0;
        memset(Head, -1, sizeof(Head));
        for(int i=1; i<=n; i++)
            dist[i] = INF;

        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( spfa() )///存在负环；
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}