World Exhibition Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 931 Accepted Submission(s): 468 Problem Description Nowadays, many people want to go to Shanghai to visit 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. Input First line: An integer T represents the case of 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. Output For each line: A single integer. If no line-up is 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. Sample Input 1 4 2 1 1 3 8 2 4 15 2 3 4 Sample Output 19 Author alpc20 Source 2010 ACM-ICPC Multi-University Training Contest（15）——Host by NUDT Recommend zhouzeyong 解題思路：基礎的差分約束，直接列不等式，然後用SPFA判斷即可，如果出現負權環即是不等式無解，如果到第n個人的最短距離一直沒有更新就說明第n個人與第1個人在圖上是不連通的，就是第n個人與第1個人的距離是任意長度，有更新的話就是有最短距離了，輸出即可。 [cpp] #include<cstdio> #include<cstring> #include<queue> #define Inf 0x7ffffff using namespace std; int head[1005],k; struct { int s; int e; int w; int next; }edge[20005]; void Add(int s,int e,int w) { edge[k].s=s; edge[k].e=e; edge[k].w=w; edge[k].next=head[s]; head[s]=k++; } int SPFA(int s) { int i,st,ed,dis[1005],cnt[1005]; bool visit[1005]; memset(visit,false,sizeof(visit)); memset(cnt,0,sizeof(cnt)); for(i=1;i<=s;i++) dis[i]=Inf; queue<int> x; x.push(1); dis[1]=0; cnt[1]++; visit[1]=true; while(!x.empty()) { st=x.front(); x.pop(); for(i=head[st];i!=-1;i=edge[i].next) { ed=edge[i].e; if(dis[ed]>dis[st]+edge[i].w) { dis[ed]=dis[st]+edge[i].w; if(!visit[ed]) { visit[ed]=true; x.push(ed); cnt[ed]++; if(cnt[ed]>s-1) return -1; } } } visit[st]=false; } if(dis[s]==Inf) return -2; else return dis[s]; } int main() { int i,t,n,x,y,a,b,c,ans; scanf("%d",&t); while(t--) { k=0; memset(head,-1,sizeof(head)); scanf("%d%d%d",&n,&x,&y); for(i=0;i<x;i++) { scanf("%d%d%d",&a,&b,&c); Add(a,b,c); } www.2cto.com for(i=0;i<y;i++) { scanf("%d%d%d",&a,&b,&c); Add(b,a,-c); } ans=SPFA(n); printf("%d\n",ans); } return 0; }
