poj3169----查分约束系统的应用

最新推荐文章于 2020-08-08 20:23:51 发布

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最新推荐文章于 2020-08-08 20:23:51 发布

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分类专栏：算法文章标签： poj 查分约束系统最短路径

本文链接：https://blog.youkuaiyun.com/lixinlei1994/article/details/75199717

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本文探讨了一种布局问题，即如何安排一群有特定距离偏好的牛的位置，以满足它们之间的最大和最小距离限制。通过将问题转化为图的最短路径问题，利用差分约束系统进行求解，并给出了具体的实现代码。

Layout
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 11702 Accepted: 5603
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.

解题思路：这里使用查分约束系统，将一个线性规划的问题转换成图的最短路径问题，查分约束系统详情可查看http://blog.youkuaiyun.com/xuezhongfenfei/article/details/8685313

#include <iostream>
#include <memory>
#include <limits.h>
#include <algorithm>
using namespace std;
const int MAX_N = 1000;
const int MAX_ML = 10000;
const int MAX_MD = 10000;
int d[MAX_N];

int AL[MAX_ML];
int BL[MAX_ML];
int DL[MAX_ML];

int AD[MAX_MD];
int BD[MAX_MD];
int DD[MAX_MD];
int main() {
    int N,ML,MD;
    cin>>N>>ML>>MD;
    fill(d,d + N,INT_MAX);
    for(int i = 0;i < ML;i++){
        cin>>AL[i]>>BL[i]>>DL[i];
    }
    for(int i = 0;i < MD;i++){
        cin>>AD[i]>>BD[i]>>DD[i];
    }
    d[0] = 0;
    //使用Bellman_Ford求解
    for(int k = 0;k < N - 1;k++){
        for(int i = 0;i + 1 < N;i++){
            //d[i]<=d[i+!],从i+1向i引一条权值为0的边
            if(d[i + 1] < INT_MAX){
                d[i] = min(d[i],d[i + 1]);
            }
        }
        for(int i = 0;i < ML;i++){
            //d[AL[i] - 1] + DL[i]>=d[BL[i] - 1],从AL[i] - 1向BL[i] - 1引一条权值为DL[i]的边
            if(d[AL[i] - 1] < INT_MAX){
                d[BL[i] - 1] = min(d[BL[i] - 1],d[AL[i] - 1] + DL[i]);
            }
        }
        for(int i = 0;i < MD;i++){
            //d[AD[i] - 1] + DL[i]<=d[BD[i] - 1],从BD[i] - 1向AD[i] - 1引一条权值为-DD[i]的边
            if(d[BD[i] - 1] < INT_MAX){
                d[AD[i] - 1] = min(d[AD[i] - 1],d[BD[i] - 1] - DD[i]);
            }
        }
    }
    int res = d[N - 1];
    //检测是否有负环
    for(int i = 0;i + 1 < N;i++){
        if(d[i + 1] < INT_MAX && d[i] > d[i + 1]){
            res = -1;
        }
    }
    for(int i = 0;i < ML;i++){
        if(d[AL[i] - 1] < INT_MAX && d[BL[i] - 1] > d[AL[i] - 1] + DL[i]){
            res = -1;
        }
    }
    for(int i = 0;i < MD;i++){
        if(d[BD[i] - 1] < INT_MAX && d[AD[i] - 1] > d[BD[i] - 1] - DD[i]){
            res = -1;
        }
    }

    if(res == INT_MAX){
        //解为无穷大
        res = -2;
    }
    cout<<res<<endl;
    return 0;
}