POJ 3169 Layout （差分约束系统）

Layout

题目链接：

Rhttp://acm.hust.edu.cn/vjudge/contest/122685#problem/S

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.

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##Input
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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.
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##Output
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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.
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##Sample Input
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4 2 1
1 3 10
2 4 20
2 3 3
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##Sample Output
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27
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##Hint
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</big>





<br/>
##题意：
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以两种形式给出若干组大小关系：
A和B的权值差最多是D.
A和B的权值差最少是D.
求#1和#N之间的最大权值差.
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<br/>
##题解：
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差分约束系统. 还需强化练习. 挖坑待填.
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<br/>
##代码：
``` cpp
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <map>
#include <set>
#include <vector>
#define LL long long
#define eps 1e-8
#define maxn 31000
#define mod 1000000007
#define inf 0x3f3f3f3f
#define IN freopen("in.txt","r",stdin);
using namespace std;

int n,m1,m2;
int u[maxn],v[maxn],w[maxn];
int first[maxn], _next[maxn], edges;
int dis[maxn];

void add_edge(int s, int t, int ww) {
    u[edges] = s; v[edges] = t; w[edges] = ww;
    _next[edges] = first[s];
    first[s] = edges++;
}

int bell_man(int s, int t) {
    for(int i=1; i<=n; i++) dis[i]=inf; dis[s] = 0;
    for(int i=1; i<=n; i++) {
        for(int e=0; e<edges; e++) if(dis[v[e]] > dis[u[e]]+w[e]){
            dis[v[e]] = dis[u[e]]+w[e];
            if(i == n) return -1;
        }
     }
     if(dis[t] == inf) return -2;
     return dis[t];
}

int main(void)
{
    //IN;

    while(scanf("%d %d %d", &n, &m1, &m2) != EOF)
    {
        memset(first, -1, sizeof(first)); edges = 0;

        while(m1--) {
            int u,v,w; scanf("%d %d %d", &u, &v, &w);
            add_edge(u, v, w);
        }
        while(m2--) {
            int u,v,w; scanf("%d %d %d", &u, &v, &w);
            add_edge(v, u, -w);
        }
        for(int i=1; i<n; i++)
            add_edge(i+1, i, 0);

        int ans = bell_man(1, n);
        printf("%d\n", ans);
    }

    return 0;
}