# World Exhibition

**Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1974 Accepted Submission(s): 979**

the World Exhibition. So there are always a lot of people who are standing along

a straight line waiting for entering. Assume that there are N (2 <= N <=

1,000) people numbered 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.

test.

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.

possible, output -1. If person 1 and N can be arbitrarily far apart, output -2.

Otherwise output the greatest possible distance between person 1 and N.

4 2 1

1 3 8

2 4 15

2 3 4

ACM-ICPC Multi-University Training Contest（15）——Host by NUDT

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#include<queue> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #define MAXN 1000010 using namespace std; queue<int>que; int n,x,y,t,tot; int vis[MAXN],dis[MAXN],cnt[MAXN]; int to[MAXN],net[MAXN],cap[MAXN],head[MAXN]; void add(int u,int v,int w){ to[++tot]=v;cap[tot]=w;net[tot]=head[u];head[u]=tot; } void spfa(){ memset(cnt,0,sizeof(cnt)); memset(vis,0,sizeof(vis)); memset(dis,0x7f,sizeof(dis)); while(!que.empty()) que.pop(); dis[1]=0;vis[1]=1; cnt[1]=1;que.push(1); while(!que.empty()){ int now=que.front(); que.pop();vis[now]=0; for(int i=head[now];i;i=net[i]) if(dis[to[i]]>dis[now]+cap[i]){ dis[to[i]]=dis[now]+cap[i]; if(!vis[to[i]]){ if(++cnt[to[i]]>n){ printf("-1\n");return ; } vis[to[i]]=1; que.push(to[i]); } } } if(dis[n]==2139062143) printf("-2\n"); else printf("%d\n",dis[n]); } int main(){ scanf("%d",&t); while(t--){ scanf("%d%d%d",&n,&x,&y); for(int i=1;i<=x;i++){ int a,b,d; scanf("%d%d%d",&a,&b,&d); add(a,b,d); } for(int i=1;i<=y;i++){ int a,b,d; scanf("%d%d%d",&a,&b,&d); add(b,a,-d); } for(int i=1;i<=n;i++) add(i,i-1,0); spfa();tot=0;memset(head,0,sizeof(head)); } } /* 2 4 2 1 1 3 8 2 4 15 2 3 4 4 2 1 1 3 10 2 4 20 2 3 3 */