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 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
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 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.
题意分析:大体意思就是有f个农场,有n个点,有m条双向路,有w个单向虫洞连接着个点,构造最短路径,看是否存在负环,就是通过虫洞倒退时间看能够存在负数环,来使一个点的出发时间小于到达时间,用bellman就可以过
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
#define ll long long
const int maxn=5202;
struct node{
int u,v,w;
};
node edge[maxn];
int n,m,p,dist[maxn],k;
void addedge(int u,int v,int w)
{
edge[k].u=u;
edge[k].v=v;
edge[k].w=w;
k++;
}
bool bellman_ford()
{
int i,j;
memset(dist,0,sizeof(dist));
for (i=1;i<=n;i++)
for (j=0;j<k;j++)
if (dist[edge[j].u]+edge[j].w<dist[edge[j].v])
dist[edge[j].v]=dist[edge[j].u]+edge[j].w;
for (j=0;j<k;j++)
if (dist[edge[j].u]+edge[j].w<dist[edge[j].v])
return 0;
return 1;
}
int main()
{
int t;
int i,u,v,w;
scanf("%d",&t);
while (t--)
{
k=0;
scanf("%d%d%d",&n,&m,&p);
for (i=0;i<m;i++)
{
scanf("%d%d%d",&u,&v,&w);
addedge(u,v,w);
addedge(v,u,w);
}
for (i=0;i<p;i++)
{
scanf("%d%d%d",&u,&v,&w);
addedge(u,v,-w);
}
if (bellman_ford()==0)
printf("YES\n");
else
printf("NO\n");
}
return 0;
}
spfa算法一样
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
#define ll long long
const int INF=1<<30;
const int maxn=5202;
struct node{
int v,w,next;
};
node edge[maxn];
int m,n,p,k,dist[maxn],head[maxn],cnt[maxn];
bool vis[maxn];
void addedge(int u,int v,int w)
{
edge[k].v=v;
edge[k].w=w;
edge[k].next=head[u];
head[u]=k++;
}
bool spfa()
{
int i;
memset(vis,0,sizeof(vis));
memset(cnt,0,sizeof(cnt));
for (i=0;i<k;i++)
dist[i]=INF;
cnt[1]++;
dist[1]=0;
vis[1]=1;
queue<int>q;
q.push(1);
while (!q.empty())
{
int x=q.front();
q.pop();
vis[x]=0;
for (int e=head[x];e!=-1;e=edge[e].next)
{
if (dist[edge[e].v]>dist[x]+edge[e].w)
{
dist[edge[e].v]=dist[x]+edge[e].w;
if (!vis[edge[e].v])
{
q.push(edge[e].v);
vis[edge[e].v]=1;
cnt[edge[e].v]++;
if (cnt[edge[e].v]>=n)
return 0;
}
}
}
}
return 1;
}
int main()
{
int t,i,j,u,v,w;
scanf("%d",&t);
while (t--)
{
k=0;
scanf("%d%d%d",&n,&m,&p);
memset(head,-1,sizeof(head));
for (i=0;i<m;i++)
{
scanf("%d%d%d",&u,&v,&w);
addedge(u,v,w);
addedge(v,u,w);
}
for (i=0;i<p;i++)
{
scanf("%d%d%d",&u,&v,&w);
addedge(u,v,-w);
}
if (spfa())
printf("NO\n");
else
printf("YES\n");
}
return 0;
}