Layout
| Time Limit: 1000MS |
|
Memory Limit: 65536K |
| Total Submissions: 6582 |
|
Accepted: 3182 |
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.
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.
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 1
1 3 10
2 4 20
2 3 3
Sample Output
27
裸的差分约束
主要是判断-1和-2的情况
当1到n的距离为无穷大的时候，说明可以选1到n的所有数
当存在负环的时候，则说明1到n的距离为无穷小，说明选不了任何点嘛！
#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <string>
#include <algorithm>
#include <queue>
using namespace std;
const int maxn = 1000+10;
const int INF = 1e9;
vector<pair<int,int> > g[maxn];
int dist[maxn];
int cnt[maxn];
queue<int>que;
bool inQue[maxn];
int src = 1;
int n,ml,md;
bool spfa(){
memset(inQue,0,sizeof inQue);
memset(cnt,0,sizeof cnt);
for(int i = 1; i <= n; i++) dist[i] = INF;
dist[src]= 0;
que.push(src);
inQue[src] = 1;
while(!que.empty()){
int u = que.front();
que.pop();
for(int i = 0; i < g[u].size(); i++){
if(dist[g[u][i].first]>g[u][i].second+dist[u]){
dist[g[u][i].first] = g[u][i].second+dist[u];
if(!inQue[g[u][i].first]){
cnt[g[u][i].first]++;
if(cnt[g[u][i].first] > n)
return false;
inQue[g[u][i].first] = 1;
que.push(g[u][i].first);
}
}
}
inQue[u] = 0;
}
return true;
}
int main(){
while(cin >> n >> ml >> md){
for(int i = 1; i <= n; i++){
g[i].clear();
}
int a,b,d;
while(ml--){
scanf("%d%d%d",&a,&b,&d);
g[a].push_back(make_pair(b,d));
}
while(md--){
scanf("%d%d%d",&a,&b,&d);
g[b].push_back(make_pair(a,-d));
}
if(!spfa()) cout<<-1<<endl;
else{
if(dist[n]==INF){
cout<<-2<<endl;
}else{
cout<<dist[n]<<endl;
}
}
}
return 0;
}
POJ3169-Layout,布布扣,bubuko.com
POJ3169-Layout
原文：http://blog.csdn.net/mowayao/article/details/32179481
