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 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.

</big>

##Input
<big>
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.
</big>

##Output
<big>
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.
</big>

##Sample Input
<big>
4 2 1
1 3 10
2 4 20
2 3 3
</big>

##Sample Output
<big>
27
</big>

##Hint
<big>
</big>

<br/>
##题意：
<big>
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.
</big>

<br/>
##题解：
<big>
差分约束系统. 还需强化练习. 挖坑待填.
</big>

<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;
}