| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 32393 | Accepted: 11771 |
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.
Source
题意：John有个n个农场，所有农场之间有m条路，每条路都有通过所需的时间。有一天，他在农场转悠，竟然发现了w个虫洞（估计这孩子小学物理学的不错~~~），每个虫洞然都有倒退的时间，他就想利用这些虫洞回到以前。问他能不能实现这个神奇之旅。
解析：就是单纯的判断是否有负环，因为如果有负环的话，就代表着回到原点的时间在出发之前，这就会到的以前。不过要特别注意路是双向的，但是虫洞确是单向的！！！级的数组一定要开的足够大，不然就RE，因为路是双向的，所以存路就需要2*MAX_E，再加上MAX_W个虫洞，总共就需要开到 2*MAX_E + MAX_W 才可以。
PS：bellman_ford算法判断负环有两种方法，但是两者的实质都是一样的，不用纠结的，如果直接就只是判断有无负环，还是版本二好写一点。
AC代码：
判负环版本一：
#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
#define INF 1234567
#define MAX_V 505
#define MAX_E 5205
struct edge{ int s, e, w; };
edge es[MAX_E];
int d[MAX_V];
int E, V;
bool Bellman_ford(int s){
for(int i=1; i<=V; i++) d[i] = (i==s) ? 0 : INF;
for(int i=0; i<V-1; i++){
int flag = 0;
for(int j=0; j<E; j++){
edge now = es[j];
if(d[now.e] > d[now.s] + now.w){
d[now.e] = d[now.s] + now.w;
flag = 1;
}
}
if(flag == 0) break;
}
for(int i=0; i<E; i++){ //判负环
edge now = es[i];
if(d[now.e] > d[now.s] + now.w)
return true;
}
return false;
}
int main(){
// freopen("in.txt", "r", stdin);
int t, n, m, w, s, e, time;
scanf("%d", &t);
while(t--){
scanf("%d%d%d", &n, &m, &w);
V = n; //顶点数
E = 0; //边数
for(int i=0; i<m; i++){ //路是双向的
scanf("%d%d%d", &s, &e, &time);
es[E].s = s;
es[E].e = e;
es[E++].w = time;
es[E].s = e;
es[E].e = s;
es[E++].w = time;
}
for(int i=0; i<w; i++){ //虫洞是单向！！！
scanf("%d%d%d", &s, &e, &time);
es[E].s = s;
es[E].e = e;
es[E++].w = -time;
}
printf("%s\n", flag ? "YES" : "NO");
}
return 0;
}
判负环版本二：
#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
#define INF 1234567
#define MAX_V 505
#define MAX_E 5205
struct edge{ int s, e, w; };
edge es[MAX_E];
int d[MAX_V];
int E, V;
bool Bellman_ford(int s){
for(int i=1; i<=V; i++) d[i] = (i==s) ? 0 : INF;
for(int i=0; i<V; i++){
for(int j=0; j<E; j++){
edge now = es[j];
if(d[now.e] > d[now.s] + now.w){
d[now.e] = d[now.s] + now.w;
if(i == V-1) return true;
}
}
}
return false;
}
int main(){
// freopen("in.txt", "r", stdin);
int t, n, m, w, s, e, time;
scanf("%d", &t);
while(t--){
scanf("%d%d%d", &n, &m, &w);
V = n;
E = 0;
for(int i=0; i<m; i++){ //路是双向的
scanf("%d%d%d", &s, &e, &time);
es[E].s = s;
es[E].e = e;
es[E++].w = time;
es[E].s = e;
es[E].e = s;
es[E++].w = time;
}
for(int i=0; i<w; i++){ //虫洞是单向！！！
scanf("%d%d%d", &s, &e, &time);
es[E].s = s;
es[E].e = e;
es[E++].w = -time;
}
printf("%s\n", Bellman_ford(1) ? "YES" : "NO");
}
return 0;
}
小编福利：如果想求出来所有的负环的话，只需要把将版本二中的d数组全部初始化为0即可~~~
