poj_3259 Wormholes（bollman-ford / SPFA)

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 45517		Accepted: 16797

Description

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.

裸Bollman-ford，找负环。也可以用SPFA。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <stack>
#include <bitset>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <algorithm>
#define FOP freopen("data.txt","r",stdin)
#define inf 0x3f3f3f3f
#define maxn 3010
#define mod 1000000007
#define PI acos(-1.0)
#define LL long long
using namespace std;

struct Edge
{
    int a, b;
    int c;
}edge[maxn * 2];

int dis[maxn];
int n, m, w;

bool Bollman_ford()
{
    for(int i = 1; i <= n; i++)
    {
        if(i == 1) dis[i] = 0;
        else dis[i] = inf;
    }

    for(int i = 1; i <= n - 1; i++)
    {
        for(int j = 1; j <= m * 2 + w; j++)
        {
            int a = edge[j].a, b = edge[j].b;
            int t = dis[a] + edge[j].c;
            if(dis[b] > t) dis[b] = t;
        }
    }

    for(int i = 1; i <= m * 2 + w; i++)
    {
        int a = edge[i].a, b = edge[i].b;
        int t = dis[a] + edge[i].c;
        if(dis[b] > t) return true;
    }
    return false;
}

int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        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);
            edge[i].a = s, edge[i].b = e, edge[i].c = t;
            edge[i+m].a = e, edge[i+m].b = s, edge[i+m].c = t;
        }
        for(int i = 1; i <= w; i++)
        {
            scanf("%d%d%d", &s, &e, &t);
            edge[i+2*m].a = s, edge[i+2*m].b = e, edge[i+2*m].c = -t;
        }
        if(Bollman_ford()) printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}

SPFA 分了bfs和dfs两种实现

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <stack>
#include <bitset>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <algorithm>
#define FOP freopen("data.txt","r",stdin)
#define inf 0x3f3f3f3f
#define maxn 3010
#define mod 1000000007
#define PI acos(-1.0)
#define LL long long
using namespace std;

struct Edge
{
    int b, c, next;
} edge[maxn * 2];

int head[maxn];
int dis[maxn];
bool vis[maxn];
int c[maxn];
int cot;
int n, m, w;

bool SPFA_bfs(int s)
{
    queue<int> Q;
    Q.push(s), vis[s] = true, c[s]++, dis[s] = 0;

    while(!Q.empty())
    {
        int a = Q.front();
        Q.pop(), vis[a] = false;
        for(int i = head[a]; i != -1; i = edge[i].next)
        {
            int b = edge[i].b;
            if(dis[b] > dis[a] + edge[i].c)
            {
                dis[b] = dis[a] + edge[i].c;
                if(!vis[b])
                {
                    Q.push(b), vis[b] = true, c[b]++;
                    if(c[b] > n - 1)return true;
                }
            }
        }
    }
    return false;
}

bool SPFA_dfs(int s)
{
    vis[s] = true;
    for(int i = head[s]; i != -1; i = edge[i].next)
    {
        int b = edge[i].b;
        if(dis[b] > dis[s] + edge[i].c)
        {
            dis[b] = dis[s] + edge[i].c;
            if(!vis[b])
            {
                if(SPFA_dfs(b)) return true;
            }
            else return true;
        }
    }
    vis[s] = false;
    return false;
}

int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d%d%d", &n, &m, &w);
        cot = 0;
        memset(head, -1, sizeof(head));

        memset(vis, false, sizeof(vis));
        memset(c, 0, sizeof(c));
        for(int i = 1; i <= n; i++) dis[i] = inf;
        int s, e, t;
        for(int i = 1; i <= m; i++)
        {
            scanf("%d%d%d", &s, &e, &t);
            edge[cot].b = e, edge[cot].c = t, edge[cot].next = head[s], head[s] = cot++;
            edge[cot].b = s, edge[cot].c = t, edge[cot].next = head[e], head[e] = cot++;
        }
        for(int i = 1; i <= w; i++)
        {
            scanf("%d%d%d", &s, &e, &t);
            edge[cot].b = e, edge[cot].c = -t, edge[cot].next = head[s], head[s] = cot++;
        }
        dis[1] = 0;
        if(SPFA_dfs(1)) printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}