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.
练习一下SPFA。
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。
SPFA:判断有没有入队次数超过N的点。Bellman_Ford就直接判断了。
#include
#include
#include
#include
#include
#include
using namespace std; const int INF=0x3f3f3f3f; const int maxn=3000*2; int end[maxn],cost[maxn]; int next[maxn],cnt[maxn],head[maxn]; int t,n,m,w,e; bool flag; int d[maxn],visit[maxn]; void add(int u,int v,int w) { end[e]=v; cost[e]=w; next[e]=head[u]; head[u]=e++; } void SPFA() { queue
q; memset(visit,0,sizeof(visit)); memset(cnt,0,sizeof(cnt)); memset(d,INF,sizeof(d)); visit[1]=1; cnt[1]++; d[1]=0; q.push(1); while(!q.empty()) { int uu=q.front(); q.pop(); visit[uu]=0; for(int i=head[uu];i!=-1;i=next[i]) { int vv=end[i]; int ww=cost[i]; if(d[vv]>d[uu]+ww) { d[vv]=d[uu]+ww; if(!visit[vv]) { visit[vv]=1; q.push(vv); if(++cnt[vv]>=n) { flag=false; return ; } } } } } return ; } int main() { int x,y,z; scanf("%d",&t); while(t--) { scanf("%d%d%d",&n,&m,&w); e=0; memset(head,-1,sizeof(head)); for(int i=0;i
Bellman_Ford：
#include
#include
#include
#include
using namespace std; #define MAXN 3000*2 #define INF 0xFFFFFFF int t , n , m, w; int dis[MAXN]; struct Edge{ int x; int y; int value; }e[MAXN]; bool judge(){ for(int i = 0 ; i < m*2+w ; i++){ if(dis[e[i].y] > dis[e[i].x] + e[i].value) return false; } return true; } void Bellman_Ford(){ dis[1] = 0; for(int i = 2 ; i <= n ; i++) dis[i] = INF; for(int i = 1 ; i <= n ; i++){ for(int j = 0 ; j < m*2+w; j++){ if(dis[e[j].y] > dis[e[j].x] + e[j].value) dis[e[j].y] = dis[e[j].x] + e[j].value; } } if(judge()) printf("NO\n"); else printf("YES\n"); } int main() { int uu,vv,ww,i; scanf("%d",&t); while(t--) { scanf("%d%d%d",&n,&m,&w); for(i=0;i