Bellman-Ford || SPFA :Wormholes

E - Wormholes

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

Submit Status

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.

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;
}