Bellman-Ford || SPFA :Wormholes

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Farmer John 在探索多个农场时发现了一系列奇特的虫洞，这些虫洞能够将他送到进入之前的时间点。通过详细的农场地图，FJ 设计了一套算法来确定是否有可能回到出发点之前的时间点，实现时间旅行的梦想。本文详细介绍了这个问题的背景、输入输出格式、以及解决策略，包括使用 Floyd 算法、Bellman-Ford 算法和 SPFA 算法进行求解的过程。

E - Wormholes

Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

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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

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.

这题第一印象得用Floyd做，死活过不了

转战Bellman-Ford和SPFA，数据有点弱，只用把第一点当做源点即可

还有就是一定要记得写好每条初始化语句，wa了无数次才查出来...

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>

using namespace std;

const int INF = 1<<31-1;

int n, m, w;
int d[550];
int cnt;

struct edge{
    int u, v, w;
}e[6000];

bool bellman(){
    int i, j, k;
    for(i = 1; i <= n; i ++){
        d[i] = INF;
    }
    d[1] = 0;
    for(k = 0; k < n-1; k ++){
        int flag = 1;// 判断是否松弛完毕
        for(i = 1; i <= cnt; i ++){
            int x = e[i].u;
            int y = e[i].v;
            if(d[x] != INF)
                if(d[y] > d[x] + e[i].w){
                    d[y] = d[x] + e[i].w;
                    flag = 0;
                }
        }
        if(flag)
            break;
    }

    for(i = 1; i <= cnt; i ++){
        if(d[e[i].v] > d[e[i].u] + e[i].w)
            return true; //存在负权回路
    }
    return false;
}

int main(){
    int f;
    int i;
    cin >> f;
    while(f -- ){
        cnt = 0;
        scanf("%d %d %d", &n, &m, &w);
        for(i = 1; i <= m; i ++){
            cnt ++;
            scanf("%d %d %d", &e[cnt].u, &e[cnt].v, &e[cnt].w);
            cnt ++;
            e[cnt].u = e[cnt-1].v;
            e[cnt].v = e[cnt-1].u;
            e[cnt].w = e[cnt-1].w;
        }
        for(i = 1; i <= w; i ++){
            cnt ++;
            scanf("%d %d %d", &e[cnt].u, &e[cnt].v, &e[cnt].w);
            e[cnt].w = -e[cnt].w;
        }
        if(bellman())
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <queue>

using namespace std;

const int INF = 1<<31-1;

int n, m, ww;
int d[55000];
int cnt;

struct node{
    int to;
    int next;//每次插边的时候是插在已插的边前面，所以已插的边是新边的next
    int weight;
}e[60000];

int lastshow[40000], num[40000];
bool inqueue[40000];
queue<int>q;

void insert(int a, int b, int w){
    e[++cnt].to = b;
    e[cnt].weight = w;
    e[cnt].next = lastshow[a];
    lastshow[a] = cnt;//lastshow记录一个点上次作为起点的边的序号，所以一条边的next是它的起点的lastshow值
}

bool spfa(){
    q.push(1);
    num[1] ++;
    while(!q.empty()){
        int x = q.front();
        q.pop();
        inqueue[x] = false;
        int id = lastshow[x];
        while(id != -1){
            if(d[x] < INF && d[e[id].to] > e[id].weight + d[x]){
                d[e[id].to] = e[id].weight + d[x];
                if(!inqueue[e[id].to]){
                    inqueue[e[id].to] = true;
                    q.push(e[id].to);
                }
                num[id]++;
                if( num[id] > n )//如果入队的次数超过总数，说明存在回路
                    return true;
            }
            id = e[id].next;
        }
    }
    return false;
}

int main(){
    int f;
    int i;
    int a, b, w;
    cin >> f;
    while(f -- ){
        cnt = 0;
        scanf("%d %d %d", &n, &m, &ww);
        memset(lastshow,-1,sizeof(lastshow));
        memset(inqueue, false, sizeof(inqueue));
        memset(num, 0, sizeof(num));
        for(i = 1; i <= n; ++ i)
            d[i]=INF;
        d[1]=0;
        cnt=0;
        while(!q.empty())
        {
            q.pop();
        }
        for(i = 1; i <= m; i ++){
            scanf("%d %d %d", &a, &b, &w);
            insert(a, b, w);
            insert(b, a, w);
        }
        for(i = 1; i <= ww; i ++){
            scanf("%d %d %d", &a, &b, &w);
            insert(a, b, -w);
        }
        if(spfa())
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}