Wormholes poj 3259 （bellman-floyd+spfa)

问题描述

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 toF (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.

输入

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 fromS to
E that also moves the traveler back T seconds.

输出

Lines 1.. F: For each farm, output "YES" if FJ can achieve his goal, otherwise output "NO" (do not include the quotes).

样例输入

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

样例输出

NO
YES

裸的bellman。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <map>
using namespace std;

const int INF=999999999;

int dist[502];

struct Edge
{
int u,v,cost;
}edge[6000];

int n,m,w;int top;

bool bellman()
{

for(int i=1;i<=n;i++)
dist[i]=INF;
dist[1]=0;
for(int i=0;i<n-1;i++)
for(int j=1;j<=top;j++)
{
int u=edge[j].u;
int v=edge[j].v;
if(dist[v]>dist[u]+edge[j].cost)
{
if(i==n-2)
return false;
dist[v]=dist[u]+edge[j].cost;
}

}
return true;

}

void add(int a,int b,int c)
{
edge[top].u=a;
edge[top].v=b;
edge[top++].cost=c;
}

int main()
{
int f;
scanf("%d",&f);
while(f--)
{
top=1;
scanf("%d %d %d",&n,&m,&w);
for(int i=1;i<=m;i++)
{
int a,b,c;
scanf("%d %d %d",&a,&b,&c);
add(a,b,c);
add(b,a,c);
}
for(int i=1;i<=w;i++)
{
int a,b,c;
scanf("%d %d %d",&a,&b,&c);
add(a,b,-c);
}
if(bellman())printf("NO\n");
else
printf("YES\n");
}
}

spfa

#include <iostream>
#include <cstdio>
#include <cstring>
#include <map>
#include <vector>
#include <queue>
using namespace std;

const int INF=999999999;

int n,m,w;

struct Edge
{
int v,w,next;

}edge[6000];

int dis[501],inq[501],head[501],used[501];
int top;

void init()
{
for(int i=0;i<=n;i++)
{
dis[i]=INF;
inq[i]=0;
head[i]=-1;
used[i]=0;
}
dis[1]=0;
top=1;
}

void add(int a,int b,int c)
{
edge[top].v=b;
edge[top].w=c;
edge[top].next=head[a];
head[a]=top++;
}

bool spfa()
{
queue<int> que;
que.push(1);
used[1]++;
while(!que.empty())
{
int u=que.front();que.pop();
inq[u]=0;
for(int s=head[u];s!=-1;s=edge[s].next)
{
if(dis[edge[s].v]>dis[u]+edge[s].w)
{
dis[edge[s].v]=dis[u]+edge[s].w;
if(!inq[edge[s].v])
{
inq[edge[s].v]=1;
que.push(edge[s].v);
if(++used[edge[s].v]>n-1)return false;
que.push(edge[s].v);
}
}

}
}
return true;
}

int main()
{
int f;
scanf("%d",&f);
while(f--)
{
scanf("%d %d %d",&n,&m,&w);
init(); int a,b,c;

for(int i=1;i<=m;i++)
{

scanf("%d %d %d",&a,&b,&c);
add(a,b,c);
add(b,a,c);
}
for(int i=1;i<=w;i++)
{

scanf("%d %d %d",&a,&b,&c);
add(a,b,-c);
}
if(!spfa())printf("YES\n");
else
printf("NO\n");

}
}