Wormholes
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 65655 | Accepted: 24488 |
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 Tseconds 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 8Sample Output
NO
YESHint
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
问题链接：POJ3259 Wormholes
问题描述：FJ有F个农场，每个农庄里有N块地，有M条双向边：(S,E,T)表示S和E之间有一条双向边，需要T时间才能走完。
还有W个虫洞:(S,E,T)表示从S到E有一条单向边，从S到E会使得时间倒退T时间。问在每个农场里FJ能不能回到比开始还早的时间，如果可以输出YES，否则输出NO。
解题思路：实际上是求是否存在负环。
AC的C++程序：
Floyd算法判断是否有负环
#include<iostream>
using namespace std;
const int N=505;
const int INF=10000000;
int dp[N][N];
//floyd算法判断是否存在负环
bool floyd(int n)
{
for(int k=1;k<=n;k++)
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++)
if(dp[i][j]>dp[i][k]+dp[k][j])
dp[i][j]=dp[i][k]+dp[k][j];
if(dp[i][i]<0)
return true;
}
return false;
}
int main()
{
int F,N,M,W,a,b,t;
scanf("%d",&F);
while(F--){
scanf("%d%d%d",&N,&M,&W);
for(int i=1;i<=N;i++)
for(int j=1;j<=N;j++)
dp[i][j]=(i==j)?0:INF;
while(M--){
//双向边
scanf("%d%d%d",&a,&b,&t);
if(t<dp[a][b])
dp[a][b]=dp[b][a]=t;
}
while(W--){
//单向边
scanf("%d%d%d",&a,&b,&t);
dp[a][b]=-t;//负号表示使时间倒退-t
}
if(floyd(N))
printf("YES\n");
else
printf("NO\n");
}
return 0;
}Bellman_ford算法判断是否有负环
#include<iostream>
#include<cstring>
using namespace std;
int dist[505];
const int INF=10000000;
struct Edge{
int start,end,w;//从u到v有一条花费为w的路
}edge[6000];
//n个结点 m条边
bool Bellman_ford(int n,int m)
{
memset(dist,INF,sizeof(dist));
dist[1]=0;
//进行n-1次松弛操作
for(int i=1;i<n;i++)
for(int j=0;j<m;j++)
if(dist[edge[j].end]>dist[edge[j].start]+edge[j].w)
dist[edge[j].end]=dist[edge[j].start]+edge[j].w;
//判断是否存在负环
for(int j=0;j<m;j++)
if(dist[edge[j].end]>dist[edge[j].start]+edge[j].w)
return true;
return false;
}
int main()
{
int F,N,M,W,a,b,t;
scanf("%d",&F);
while(F--){
scanf("%d%d%d",&N,&M,&W);
int k=0;
while(M--){
//双向边
scanf("%d%d%d",&a,&b,&t);
edge[k].start=a,edge[k].end=b,edge[k].w=t;
k++;
edge[k].start=b,edge[k].end=a,edge[k].w=t;
k++;
}
while(W--){
//单向边
scanf("%d%d%d",&a,&b,&t);
edge[k].start=a,edge[k].end=b,edge[k].w=-t;
k++;
}
bool flag=Bellman_ford(N,k);
printf("%s\n",flag?"YES":"NO");
}
return 0;
}SPFA算法判断是否有负环：
#include<iostream>
#include<queue>
#include<vector>
using namespace std;
const int INF=0x3f3f3f3f;
const int N=505;
struct Edge{//边
int v,cost;//起点到终点v的花费为cost
Edge(int v,int cost):v(v),cost(cost){}
};
int dist[N];//距离
bool vis[N];//标记结点是否进队列
int cnt[N];//标记结点进队列的次数
vector<Edge>g[N];//图的邻接表表示
bool spfa(int n,int s)//n是结点的个数，s是源点
{
queue<int>q;
for(int i=0;i<=n;i++){
vis[i]=false;
cnt[i]=0;
dist[i]=INF;
}
vis[s]=true;
dist[s]=0;
cnt[s]=1;
q.push(s);
while(!q.empty()){
int u=q.front();
q.pop();
vis[u]=false;//出队列标记为false
int num=g[u].size();//以u为起点的终点个数
for(int i=0;i<num;i++){
int v=g[u][i].v;//终点编号
if(dist[u]+g[u][i].cost<dist[v]){
dist[v]=dist[u]+g[u][i].cost;
if(!vis[v]){
q.push(v);
vis[v]=true;
cnt[v]++;
if(cnt[v]>=n)
return false;
}
}
}
}
return true;
}
int main()
{
int T,n,m,w,a,b,t;
scanf("%d",&T);
while(T--){
scanf("%d%d%d",&n,&m,&w);
for(int i=0;i<=n;i++)
g[i].clear();
while(m--){
scanf("%d%d%d",&a,&b,&t);
g[a].push_back(Edge(b,t));
g[b].push_back(Edge(a,t));
}
while(w--){
scanf("%d%d%d",&a,&b,&t);
g[a].push_back(Edge(b,-t));
}
int flag=spfa(n,1);
if(flag)
printf("NO\n");
else
printf("YES\n");
}
return 0;
}