poj3259 Wormholes【最短路-bellman-负环】-白红宇

poj3259 Wormholes【最短路-bellman-负环】

阅读量：4332 次

发布时间：2019-06-06

本文共 3214 字，大约阅读时间需要 10 分钟。

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 farm descriptions follow.

Line 1 of each farm: Three space-separated integers respectively:

M, and

Lines 2..

M+1 of each farm: Three space-separated numbers (

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 (

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

23 3 11 2 21 3 42 3 13 1 33 2 11 2 32 3 43 1 8

Sample Output

NOYES

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.

题意：走一条路会花费时间走虫洞会时间倒流问能不能走回起点的时候时间倒流

思路：其实就是判断有没有环有环的话

看题的时候看了半天不知道起点到底是哪一个点

感觉现在bellman写的还挺顺手的了

一个加边的addedge函数一个判断松弛的relax函数

然后外面一个for循环遍历n-1次里面的for循环松弛每一条边

再对每一条边判断能不能松弛能就说明有环

有环其实就说明这个路径上有负的

不然干吗要不停的重复走？？？

只有可以不断减小才会形成环

WA了一发是因为加边的时候的cnt没有初始化

代码：

#include
    
     #include
     
      #include
      
       #include
       
        #include
        #include
         
          #include
          
           #include
           
            #define inf 0x3f3f3f3fusing namespace std;int f, n, m, w, cnt;struct edge{ int s, e, t;}path[5205];long long d[505];void addedge(int s, int e, int t){ path[cnt].s = s; path[cnt].e = e; path[cnt].t = t; cnt++;}bool relax(int j){ if(d[path[j].e] > d[path[j].s] + path[j].t){ d[path[j].e] = d[path[j].s] + path[j].t; return true; } return false;}bool bellman(int sec){ memset(d, inf, sizeof(d)); d[sec] = 0; for(int i = 0; i < n - 1; i++){ bool flag = false; for(int j = 0; j < cnt; j++){ if(relax(j)) flag = true; } if(!flag) return false; } for(int i = 0; i < cnt; i++){ if(relax(i)) return true; } return false;}int main(){ cin>>f; while(f--){ cin>>n>>m>>w; cnt = 0; for(int i = 0; i < m; i++){ int a, b, c; cin>>a>>b>>c; addedge(a, b, c); addedge(b, a, c); } for(int i = 0; i < w; i++){ int a, b, c; cin>>a>>b>>c; addedge(a, b, -c); } if(bellman(1)) cout<<"YES\n"; else cout<<"NO\n"; } return 0;}

转载于:https://www.cnblogs.com/wyboooo/p/9643424.html

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