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POJ 3169 Layout (差分约束)
Layout
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 10245 | Accepted: 4931 |
Description
Like everyone else, cows like to stand close to their friends when queuing for feed. FJ has N (2 <= N <= 1,000) cows numbered 1..N standing along a straight line waiting for feed. The cows are standing in the same order as they are numbered, and since they can be rather pushy, it is possible that two or more cows can line up at exactly the same location (that is, if we think of each cow as being located at some coordinate on a number line, then it is possible for two or more cows to share the same coordinate).
Some cows 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 ML (1 <= ML <= 10,000) constraints describes which cows like each other and the maximum distance by which they may be separated; a subsequent list of MD constraints (1 <= MD <= 10,000) tells which cows 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 cow 1 and cow N that satisfies the distance constraints.
Some cows 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 ML (1 <= ML <= 10,000) constraints describes which cows like each other and the maximum distance by which they may be separated; a subsequent list of MD constraints (1 <= MD <= 10,000) tells which cows 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 cow 1 and cow N that satisfies the distance constraints.
Input
Line 1: Three space-separated integers: N, ML, and MD.
Lines 2..ML+1: Each line contains three space-separated positive integers: A, B, and D, with 1 <= A < B <= N. Cows A and B must be at most D (1 <= D <= 1,000,000) apart.
Lines ML+2..ML+MD+1: Each line contains three space-separated positive integers: A, B, and D, with 1 <= A < B <= N. Cows A and B must be at least D (1 <= D <= 1,000,000) apart.
Lines 2..ML+1: Each line contains three space-separated positive integers: A, B, and D, with 1 <= A < B <= N. Cows A and B must be at most D (1 <= D <= 1,000,000) apart.
Lines ML+2..ML+MD+1: Each line contains three space-separated positive integers: A, B, and D, with 1 <= A < B <= N. Cows A and B must be at least D (1 <= D <= 1,000,000) apart.
Output
Line 1: A single integer. If no line-up is possible, output -1. If cows 1 and N can be arbitrarily far apart, output -2. Otherwise output the greatest possible distance between cows 1 and N.
Sample Input
4 2 11 3 102 4 202 3 3
Sample Output
27
Hint
Explanation of the sample:
There are 4 cows. Cows #1 and #3 must be no more than 10 units apart, cows #2 and #4 must be no more than 20 units apart, and cows #2 and #3 dislike each other and must be no fewer than 3 units apart.
The best layout, in terms of coordinates on a number line, is to put cow #1 at 0, cow #2 at 7, cow #3 at 10, and cow #4 at 27.
There are 4 cows. Cows #1 and #3 must be no more than 10 units apart, cows #2 and #4 must be no more than 20 units apart, and cows #2 and #3 dislike each other and must be no fewer than 3 units apart.
The best layout, in terms of coordinates on a number line, is to put cow #1 at 0, cow #2 at 7, cow #3 at 10, and cow #4 at 27.
/* * POJ 3169 Layout * 给你n条奶牛，其中一些两个距离不能超过D,而有些必须大于等于D,问1到n的最大距离 * * 典型的差分约束，全部化为小于等于的不等式，跑SPFA即可，由于存在负环，不能用Dijkstra。 */#include <stdio.h>#include <string.h>#include <iostream>#include <queue>using namespace std;const int MAXN = 1010;const int MAXE = 20020;const int INF = 0x3f3f3f3f;struct Edge{ int to; int w; int next;}edge[MAXE];int head[MAXN],tol;bool vis[MAXN];int cnt[MAXN];int Q[MAXN];int dis[MAXN];void init(){ tol=0; memset(head,-1,sizeof(head));}void addedge(int u,int v,int w)//加边{ edge[tol].to=v; edge[tol].w=w; edge[tol].next=head[u]; head[u]=tol++;}bool SPFA(int st,int n){ int front=0,rear=0; for(int v=1;v<=n;v++)//初始化 { if(v==st) { Q[rear++]=v; vis[v]=true; cnt[v]=1; dis[v]=0; } else { vis[v]=false; cnt[v]=0; dis[v]=INF; } } while(front!=rear) { int u=Q[front++]; vis[u]=false; if(front>=MAXN)front=0; for(int i=head[u];i!=-1;i=edge[i].next) { int v=edge[i].to; if(dis[v]>dis[u]+edge[i].w) { dis[v]=dis[u]+edge[i].w; if(!vis[v]) { vis[v]=true; Q[rear++]=v; if(rear>=MAXN)rear=0; if(++cnt[v]>n) return false; } } } } return true;}int main(){ int n,ml,md; int u,v,w; scanf("%d%d%d",&n,&ml,&md); init(); while(ml--) { scanf("%d%d%d",&u,&v,&w); addedge(u,v,w); } while(md--) { scanf("%d%d%d",&u,&v,&w); addedge(v,u,-w); } if(!SPFA(1,n)) printf("-1\n"); else if(dis[n]==INF) printf("-2\n"); else printf("%d\n",dis[n]); return 0;}
POJ 3169 Layout (差分约束)
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