World Exhibition
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 572 Accepted Submission(s): 267
1..N who are standing in the same order as they are numbered. It is possible that two or more person line up at exactly the same location in the condition that those visit it in a group.
There is something interesting. Some like each other and want to be within a certain distance of each other in line. Some really dislike each other and want to be separated by at least a certain distance. A list of X (1 <= X <= 10,000) constraints describes
which person like each other and the maximum distance by which they may be separated; a subsequent list of Y constraints (1 <= Y <= 10,000) tells which person dislike each other and the minimum distance by which they must be separated.
Your job is to compute, if possible, the maximum possible distance between person 1 and person N that satisfies the distance constraints.
The next line: Three space-separated integers: N, X, and Y.
The next X lines: Each line contains three space-separated positive integers: A, B, and C, with 1 <= A < B <= N. Person A and B must be at most C (1 <= C <= 1,000,000) apart.
The next Y lines: Each line contains three space-separated positive integers: A, B, and C, with 1 <= A < B <= C. Person A and B must be at least C (1 <= C <= 1,000,000) apart.
1 4 2 1 1 3 8 2 4 15 2 3 4
19
#include<stdio.h> #include<algorithm> #include<queue> #include<vector> #define inf 99999999 using namespace std; struct node { int v; int len; }; vector<node>map[1005]; int n,x,y; void init() { int i; for(i=0;i<1005;i++) map[i].clear(); } int spfa(int s) { int i,dis[1005]; bool used[1005]; int num[1005]; queue<int>q; for(i=0;i<1005;i++) dis[i]=inf; memset(num,0,sizeof(num)); memset(used,0,sizeof(used)); dis[s]=0; used[s]=true; q.push(s); while(!q.empty()) { int u=q.front(); q.pop(); used[u]=false; for(i=0;i<map[u].size();i++) { node p=map[u][i]; if(dis[p.v]>dis[u]+p.len) { dis[p.v]=dis[u]+p.len; if(!used[p.v]) { used[p.v]=true; num[p.v]++; if(num[p.v]>n) return -1; q.push(p.v); } } } } if(dis[n]>=inf) return -2; else return dis[n]; } int main() { int i,t,a,b,l,ans; node p; scanf("%d",&t); while(t--) { init(); scanf("%d%d%d",&n,&x,&y); for(i=1;i<=x;i++) { scanf("%d%d%d",&a,&b,&l); p.v=b; p.len=l; map[a].push_back(p); } for(i=1;i<=y;i++) { scanf("%d%d%d",&a,&b,&l); p.v=a; p.len=-l; map[b].push_back(p); } ans=spfa(1); printf("%d\n",ans); } return 0; }