Layout POJ NO.3169

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

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.

 

题解：首先记第 i 号牛的位置为d[ i ]。牛是按照顺序排列的，所以 d[ i ] <= d[ i +1 ]。其次，对于每对关系好的牛之间的最大距离限制，都有 d[ AL ]+DL >= d[ BL ]成立。

同样，对于每对关系不好的牛，都有 d[ AM ]+DM <= d[ BM ]。这些不等式的特点是所有式子两边都只出现了一个变量。实际上，图上的最短距离也可以用这样的形式

表示出来。把原来的问题和最短路问题进行比较就可以发现，两个问题就是完全一样的形式。可以通过把原来的问题的每一个约束不等式对应成图中的一条边来构图。

（1）d[ i ] <= d[ i + 1 ]变形为d[ i + 1 ] + 0 >= d[ i ]，因此从顶点 i + 1向顶点 i 连一条权值为0 的边。

（2）d[ AL ]+DL >= d[ BL ]对应从顶点AL向顶点BL连一条权值为DL的边。

（3）d[ AM ]+DM <= d[ BM ]变形为d[ BM ] - DM >= d[ AM ] ，对应从顶点BM向顶点AM连一条权值为 -DM的边。

因为图中存在负权边，所以用Bellman-Ford算法求解。复杂度O（N（N+ML+DL））-----------------------------------------------------摘自《挑战程序设计竞赛》

感受：很好很好的题，就是写不出。。。建图也是一门绝活啊，而且对最短路的几种算法并没有深刻的理解。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define INF 1e8
using namespace std;

const int maxn=10005;

int n,M,D;
int d[1005];
int AL[maxn],BL[maxn],DL[maxn];
int AM[maxn],BM[maxn],DM[maxn];

void solve(){
    for(int i=0;i<n;i++) d[i]=INF;
    d[0]=0;
    for(int k=0;k<n;k++){
        for(int i=0;i+1<n;i++) if(d[i+1]<INF) d[i]=min(d[i],d[i+1]);
        for(int i=0;i<M;i++) if(d[AL[i]-1]<INF) d[BL[i]-1]=min(d[BL[i]-1],d[AL[i]-1]+DL[i]);
        for(int i=0;i<D;i++) if(d[BM[i]-1]<INF) d[AM[i]-1]=min(d[AM[i]-1],d[BM[i]-1]-DM[i]);
    }
    int ans=d[n-1];
    if(d[0]<0) cout<<"-1"<<endl;
    else if(ans==INF) cout<<"-2"<<endl;
    else cout<<ans<<endl; 
}

int main()
{   cin>>n>>M>>D;
    for(int i=0;i<M;i++) scanf("%d%d%d",&AL[i],&BL[i],&DL[i]);
    for(int i=0;i<D;i++) scanf("%d%d%d",&AM[i],&BM[i],&DM[i]);
    solve();
}