POJ 3169 Layout

Posted 2020-09-29 mfmdaoyou

Layout

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9427		Accepted: 4517

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.

题意：

农夫养了N头牛。编号为1到N。

依照编号顺序排成一排。在他们之间，有一些牛关系比較好。所以希望彼此之间不超过一定距离，也有一些牛关系比較不好，希望彼此之间至少要满足某个距离。给出了ML个关系好的，以及MD个关系不好的，求1号牛和N号牛之间的最大距离。假设不存在不论什么一种排列方法满足条件则输出-1.无限大的情况输出-2.

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 10000 + 10;
const int INF = 100000000;

int N, ML, MD;
int AL[maxn], BL[maxn], DL[maxn];
int AD[maxn], BD[maxn], DD[maxn];
int d[1010];

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

    //用Bellman-Ford算法计算d
    for (int k = 0; k < N; k++){
        //从i+1到i的权值为0
        for (int i = 0; i + 1 < N; i++){
            if (d[i + 1] < INF)
                d[i] = min(d[i], d[i + 1]);
        }
        //从AL到BL的权值为DL
        for (int i = 0; i < ML; i++){
            if (d[AL[i] - 1] < INF){
                d[BL[i] - 1] = min(d[BL[i] - 1], d[AL[i] - 1] + DL[i]);
            }
        }
        //从BD到AD的权值为-DD
        for (int i = 0; i < MD; i++){
            if (d[BD[i] - 1] < INF){
                d[AD[i] - 1] = min(d[AD[i] - 1], d[BD[i] - 1] - DD[i]);
            }
        }
    }
    int res = d[N - 1];
    if (d[0] < 0){
        //存在负圈则无解
        res = -1;
    }
    else if (res == INF){
        res = -2;
    }
    printf("%d\n", res);
}

int main()
{
    scanf("%d%d%d", &N, &ML, &MD);
    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;
}