POJ 3259 Wormholes (Bellman)
A - Wormholes
Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
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.
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 8
Sample Output
NO YES
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include<queue>
using namespace std;
const int INF=0x3f3f3f3f;
const int MAXN=500+5;
const int MAXM=5700+5;
int u[MAXM];
int v[MAXM];
int w[MAXM];
int d[MAXN];
int MIN(int x,int y){return x>y?y:x;}
int MAX(int x,int y){return x<y?y:x;}
bool bellman(int n,int m,int t)
{
int i,j,k;
for(i=1;i<=n;i++)d[i]=INF;
d[t]=0;
for(i=1;i<n;i++)
for(j=1;j<=m;j++)
if(d[u[j]]+w[j]<d[v[j]])d[v[j]]=d[u[j]]+w[j];
for(j=1;j<=m;j++)
if(d[u[j]]+w[j]<d[v[j]])return 1;
return 0;
}
int main()
{
//freopen("123.txt","r",stdin);
int N,n,m,c;
while(~scanf("%d",&N))
{
int e,i;
while(N--)
{
scanf("%d%d%d",&n,&m,&c);
for(e=1;e<=m+c;e++)
{
scanf("%d%d%d",&u[e],&v[e],&w[e]);
if(e>m)w[e]*=-1;
}
for(e=m+c+1;e<=2*m+c;e++)
{
u[e]=v[e-m-c];
v[e]=u[e-m-c];
w[e]=w[e-m-c];
}
if(bellman(n,2*m+c,1))printf("YES\n");
else printf("NO\n");
}
}
return 0;
}优化版本：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include<queue>
using namespace std;
const int INF=0x3f3f3f3f;
const int MAXN=500+5;
const int MAXM=5700+5;
int u[MAXM];
int v[MAXM];
int w[MAXM];
int d[MAXN];
int MIN(int x,int y){return x>y?y:x;}
int MAX(int x,int y){return x<y?y:x;}
bool bellman(int n,int m,int t)
{
int i,j,k;
for(i=1;i<=n;i++)d[i]=INF;
d[t]=0;
for(i=1;i<=n;i++)
{
int flag=1;
for(j=1;j<=m;j++)
if(d[u[j]]+w[j]<d[v[j]])
{
d[v[j]]=d[u[j]]+w[j];
flag=0;
}
if(flag)break;
}
if(i==n+1)return 1;
return 0;
}
int main()
{
//freopen("123.txt","r",stdin);
int N,n,m,c;
while(~scanf("%d",&N))
{
int e,i;
while(N--)
{
scanf("%d%d%d",&n,&m,&c);
for(e=1;e<=m+c;e++)
{
scanf("%d%d%d",&u[e],&v[e],&w[e]);
if(e>m)w[e]*=-1;
}
for(e=m+c+1;e<=2*m+c;e++)
{
u[e]=v[e-m-c];
v[e]=u[e-m-c];
w[e]=w[e-m-c];
}
if(bellman(n,2*m+c,1))printf("YES\n");
else printf("NO\n");
}
}
return 0;
}
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include<queue>
using namespace std;
const int INF=0x3f3f3f3f;
const int MAXN=500+5;
const int MAXM=5700+5;
int u[MAXM];
int v[MAXM];
int w[MAXM];
int first[MAXN];
int next[MAXM];
int d[MAXN];
bool inq[MAXN];
int num[MAXN];
int MIN(int x,int y){return x>y?y:x;}
int MAX(int x,int y){return x<y?y:x;}
int cnt;
void add_dedge(int u,int v,int w,int i)
{
next[i]=first[u];
first[u]=i;
cnt++;
}
bool bellman(int n,int m,int t)
{
int i;
queue<int>q;
memset(inq,0,sizeof inq);
memset(num,0,sizeof num);
for(i=1;i<=n;i++)d[i]=INF;
d[t]=0;
q.push(t);
num[t]++;
//inq[t]=1;
while(!q.empty())
{
int e,x=q.front();q.pop();
inq[x]=0;
for(e=first[x];e!=-1;e=next[e])
{
if(d[x]+w[e]<d[v[e]])
{
d[v[e]]=d[x]+w[e];
if(!inq[v[e]])
{
q.push(v[e]);
num[v[e]]++;
if(num[v[e]]>=n)return 1;
inq[v[e]]=1;
}
}
}
}
return 0;
}
int main()
{
//freopen("123.txt","r",stdin);
int N,n,m,c;
while(~scanf("%d",&N))
{
int e,i;
while(N--)
{
cnt=1;
memset(first,-1,sizeof first);
memset(next,-1,sizeof next);
scanf("%d%d%d",&n,&m,&c);
while(m--)
{
scanf("%d%d%d",&u[cnt],&v[cnt],&w[cnt]);
add_dedge(u[cnt],v[cnt],w[cnt],cnt);
u[cnt]=v[cnt-1];
v[cnt]=u[cnt-1];
w[cnt]=w[cnt-1];
add_dedge(u[cnt],v[cnt],w[cnt],cnt);
}
while(c--)
{
scanf("%d%d%d",&u[cnt],&v[cnt],&w[cnt]);
w[cnt]*=-1;
add_dedge(u[cnt],v[cnt],w[cnt],cnt);
}
cnt--;
if(bellman(n,cnt,1))printf("YES\n");
else printf("NO\n");
}
}
return 0;
}