POJ 3169 Layout （差分约束系统 + Bellman-ford算法）

Layout

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

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.

Source

USACO 2005 December Gold

题意：一个牛舍里有N头牛，有一些牛的关系比较好，他们希望彼此不超过一定的距离。当然也有些牛关系不好，他们希望彼此超过一定的距离。有ML对牛的关系比较好，并给出每对牛的所不超过的距离D；同样，有MD对牛的关系不好，并给出每对牛的所超过的距离D。问是否有满足这样的安排方案满足所有牛的要求。若不存在，输出-1；若存在，但是牛1和牛N之间的距离可以任意大，输出-2；否则，输出牛1和牛N之间的最大距离。

解析：记第i号牛的位置是d[i]，首先，牛市按照编号顺序排的，所以有d[i] <= d[i+1]成立。其次，对于关系好的牛的最大距离限制有d[AL] + DL >= d[BL]，同样对于关系不好的牛的最小距离限制有d[AD] + DD <= d[BD]。这样，原来的问题就可以转化为在满足着三个不等式组的情况下，求解d[N] - d[1]的最大值问题。当然可以用线性规划的方法去解决。但是这道题还可以更简单的求解。这个可以抽象为一个无向图，各个牛为顶点，他们之间的好或不好的关系为边，限制的距离作为边尚德权值。这样可以将这三个不等式转化一下：d[i+1]
+ 0 >= d[i]表示从i+1到i连一条权值为0的边，d[AL] + DL >= d[BL]表示从AL到BL连一条权值为DL的边，d[BD] - DD >= d[AD]表示从BD到AD连一条权值为-DD的边。所求d[N] - d[1]的最大值，对应于d[1]到d[n]之间的最短路径。由于存在负权边，所以用bellman-ford算法。

AC代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <stack>
using namespace std;
#define INF 123456789
#define LL long long
#define MID(a, b)  a+(b-a)/2
const int maxn = 1000 + 10;
const int maxm = 10000 + 10;

int al[maxm], bl[maxm], dl[maxm];
int ad[maxm], bd[maxm], dd[maxm];
int d[maxn];
int N, ML, MD;

void solve(){
    fill(d, d+N, INF);             //初始化d[]
    d[0] = 0;

    for(int i=0; i<N; i++){
        for(int j=0; j+1<N; j++)          //检查d[j]能否通过d[j+1]松弛
            if(d[j+1] < INF) d[j] = min(d[j], d[j+1] + 0);

        for(int j=0; j<ML; j++)           //检查d[BL]能否通过d[AL]松弛
            if(d[ al[j] - 1 ] < INF) d[ bl[j] - 1 ] = min(d[ bl[j] - 1 ], d[ al[j] - 1 ] + dl[j]);

        for(int j=0; j<MD; j++)           //检查d[AD]能否通过d[BD]松弛
            if(d[ bd[j] - 1 ] < INF) d[ ad[j] - 1 ] = min(d[ ad[j] - 1 ], d[ bd[j] - 1 ] - dd[j]);
    }

    int ans = d[N-1];
    if(d[0] < 0) ans = -1;
    else if(ans == INF) ans = -2;
    printf("%d\n", ans);
}

int main(){
    #ifdef sxk
        freopen("in.txt", "r", stdin);
    #endif // sxk

    while(scanf("%d%d%d", &N, &ML, &MD)!=EOF){
        for(int i=0; i<ML; i++) scanf("%d%d%d", &al[i], &bl[i], &dl[i]);
        for(int i=0; i<MD; i++) scanf("%d%d%d", &ad[i], &bd[i], &dd[i]);
        solve();
    }
    return 0;
}