Language: Default Wormholes
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. 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 Hint For farm 1, FJ cannot travel back in time. Source |
题意：John的农场里field块地，path条路连接两块地，hole个虫洞，虫洞是一条单向路，不但会把你传送到目的地，而且时间会倒退Ts。我们的任务是知道会不会在从某块地出发后又回来，看到了离开之前的自己。
很久之前做的最短路，今天又写了一遍，复习了一下SPFA和Bellma_Ford，判断负权值回路。
代码：
//Bellman_Ford
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 550
#define MAXN 2005
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b) for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b) for(i = a; i < b; i++)
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define mem(t, v) memset ((t) , v, sizeof(t))
#define sf(n) scanf("%d", &n)
#define sff(a,b) scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf printf
#define DBG pf("Hi\n")
typedef long long ll;
using namespace std;
struct Edge
{
int u,v,w;
}edge[maxn*10];
int n,m,w,num;
int dist[maxn];
bool Bellman_Ford()
{
int i,j;
FRL(i,0,n+2)
dist[i]=INF;
dist[1]=0;
FRL(i,1,n)
{
bool flag=false;
FRL(j,0,num)
{
int u=edge[j].u;
int v=edge[j].v;
if (dist[v]>dist[u]+edge[j].w)
{
flag=true;
dist[v]=dist[u]+edge[j].w;
}
}
if (!flag) return true;
}
FRL(i,0,num)
if (dist[ edge[i].v ]>dist[ edge[i].u ]+edge[i].w)
return false;
return true;
}
int main()
{
int i,j,cas;
sf(cas);
while (cas--)
{
num=0;
int u,v,t;
sfff(n,m,w);
FRL(i,0,m)
{
sfff(u,v,t);
edge[num].u=u;
edge[num].v=v;
edge[num].w=t;
num++;
edge[num].u=v;
edge[num].v=u;
edge[num].w=t;
num++;
}
FRL(i,0,w)
{
sfff(u,v,t);
edge[num].u=u;
edge[num].v=v;
edge[num].w=-t;
num++;
}
if (Bellman_Ford())
pf("NO\n");
else
pf("YES\n");
}
return 0;
}
//SPFA
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 550
#define MAXN 2005
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b) for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b) for(i = a; i < b; i++)
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define mem(t, v) memset ((t) , v, sizeof(t))
#define sf(n) scanf("%d", &n)
#define sff(a,b) scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf printf
#define DBG pf("Hi\n")
typedef long long ll;
using namespace std;
struct Edge
{
int v,w;
int next;
}edge[maxn*10];
int n,m,w,num;
int head[maxn];
int dist[maxn];
bool inq[maxn];
int cnt[maxn];
void init()
{
num=0;
mem(head,-1);
mem(inq,false);
mem(cnt,0);
mem(dist,INF);
}
void addedge(int u,int v,int w)
{
edge[num].v=v;
edge[num].w=w;
edge[num].next=head[u];
head[u]=num++;
}
bool SPFA()
{
queue<int>Q;
inq[1]=true;
cnt[1]=1;
dist[1]=0;
while (!Q.empty()) Q.pop();
Q.push(1);
while (!Q.empty())
{
int u=Q.front(); Q.pop();
inq[u]=false;
if (cnt[u]>n)
return false;
for (int i=head[u];i!=-1;i=edge[i].next)
{
int v=edge[i].v;
if (dist[v]>dist[u]+edge[i].w)
{
dist[v]=dist[u]+edge[i].w;
if (!inq[v])
{
inq[v]=true;
cnt[v]++;
Q.push(v);
}
}
}
}
return true;
}
int main()
{
int i,j,cas;
sf(cas);
while (cas--)
{
int u,v,t;
init();
sfff(n,m,w);
FRL(i,0,m)
{
sfff(u,v,t);
addedge(u,v,t);
addedge(v,u,t);
}
FRL(i,0,w)
{
sfff(u,v,t);
addedge(u,v,-t);
}
if (SPFA())
pf("NO\n");
else
pf("YES\n");
}
return 0;
}
