Wormholes
| Time Limit: 2000MS |
Memory Limit: 65536K |
|
| Total Submissions: 61855 |
Accepted: 23054 |
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 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
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.
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
算法分析：
题意：
N块地，M条路，W个虫洞。判断有没有可以是时间倒流的路径。
分析：
开始的思想，是把有虫洞连接两点的最短路求出，然后判断虫洞时间是否大于最短路，然后一直wrong ，题意好像是遇到虫洞就必须进虫洞。
未过代码：
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
using namespace std;
const double eps = 1e-8;
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 1e5 + 10;
const int MAXT = 10000 + 10;
const int M=2005;
int n,m,f,dis[M],vis[M];
vector<int>g[M],w[M];
int spfa(int s)
{
for(int i=0;i<=n;i++)
dis[i]=INT_INF;
memset(vis,0,sizeof(vis));
queue<int>q;
q.push(s);
dis[s]=0;
vis[s]=1;
while(!q.empty())
{
int i=q.front();
q.pop();
vis[i]=0;
for(int j=0;j<g[i].size();j++)
{
int k=g[i][j];
if(dis[k]>dis[i]+w[i][j])
{
dis[k]=dis[i]+w[i][j];
if(!vis[k])
{
vis[k]=1;
q.push(k);
}
}
}
}
return 0;
}
int main()
{
int T;
cin>>T;
while(T--)
{
scanf("%d%d%d",&n,&m,&f);
memset(g,0,sizeof(g));
memset(w,0,sizeof(w));
for(int i=1;i<=m;i++)//这里要从一开始
{
int a,b,d;
scanf("%d%d%d",&a,&b,&d);
g[a].push_back(b);//g记录该点连接的点
g[b].push_back(a);
w[a].push_back(d);
w[b].push_back(d);
}
int flag=0;
for(int i=1;i<=f;i++)
{
int a,b,d;
scanf("%d%d%d",&a,&b,&d);
if(flag==0)
{
spfa(a);
if(dis[b]<d)
flag=1;
}
}
if(flag==1)
printf("YES\n");
else
printf("NO\n");
}
return 0;
}
网上搜索是一个判断负边环，
我们考虑一个节点入队的条件是什么，只有那些在前一遍松弛中改变了距离估计值的点，才可能引起他们的邻接点的距离估计值的改变。因此，用一个先进先出的队列来存放被成功松弛的顶点。同样，我们有这样的定理：“两点间如果有最短路，那么每个结点最多经过一次。也就是说，这条路不超过n-1条边。”（如果一个结点经过了两次，那么我们走了一个圈。如果这个圈的权为正，显然不划算；如果是负圈，那么最短路不存在；如果是零圈，去掉不影响最优值）。也就是说，每个点最多入队n-1次（这里比较难理解，需要仔细体会，n-1只是一种最坏情况，实际中，这样会很大程度上影响程序的效率）。
有了上面的基础，思路就很显然了，加开一个数组记录每个点入队的次数（num），然后，判断当前入队的点的入队次数，如果大于n-1，则说明存在负权回路。
代码实现：
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
using namespace std;
const double eps = 1e-8;
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 1e5 + 10;
const int MAXT = 10000 + 10;
const int M=2005;
int n,m,f,dis[M],vis[M],num[M],flag=0;
vector<int>g[M],w[M];
int spfa(int s)
{
for(int i=0;i<=n;i++)
dis[i]=INT_INF;
memset(vis,0,sizeof(vis));
memset(num,0,sizeof(num));
queue<int>q;
q.push(s);
dis[s]=0;
vis[s]=1;
num[1]++;
while(!q.empty())
{
int i=q.front();
q.pop();
vis[i]=0;
for(int j=0;j<g[i].size();j++)
{
int k=g[i][j];
if(dis[k]>dis[i]+w[i][j])
{
dis[k]=dis[i]+w[i][j];
if(!vis[k])
{
num[k]++;
vis[k]=1;
q.push(k);
if(num[k]>n) {flag=1;break; }
}
}
}
if(flag==1) break;
}
return 0;
}
int main()
{
int T;
cin>>T;
while(T--)
{
flag=0;
scanf("%d%d%d",&n,&m,&f);
memset(g,0,sizeof(g));
memset(w,0,sizeof(w));
for(int i=1;i<=m;i++)//这里要从一开始
{
int a,b,d;
scanf("%d%d%d",&a,&b,&d);
g[a].push_back(b);//g记录该点连接的点
g[b].push_back(a);
w[a].push_back(d);
w[b].push_back(d);
}
for(int i=1;i<=f;i++)
{
int a,b,d;
scanf("%d%d%d",&a,&b,&d);
g[a].push_back(b);//g记录该点连接的点
w[a].push_back(-d);
}
spfa(1);
if(flag==1)
printf("YES\n");
else
printf("NO\n");
}
return 0;
}