POJ--3169 Layout(最短路)

记录
12:36 2023-1-30

http://poj.org/problem?id=3169

reference:《挑战程序设计竞赛(第2版)》2.5.6 p111

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

最短路径问题。将距离关系等式表达出来后，发现和最短路问题的表达形式一样，所以转化为了最短路问题。需要注意的是，存在负边，所以不能使用dijkstra，要使用Bellman-Ford算法。这个问题关键的部分是要将关系等式看懂并联系到最短路上，而且使用最短路计算出的最短距离就是1号与N号之间的最大距离，因为计算的时候使用的是关系好之间的最大值，关系差之间的最小值。一些约束不等式的题可以转化为最短路问题

#include<cstdio>
#include<vector>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
#define MAX_ML 10000
#define MAX_MD 10000
#define MAX_N 10000

const int INF = 0x3f3f3f3f;

int N, ML, MD;
int AL[MAX_ML], BL[MAX_ML], DL[MAX_ML];
int AD[MAX_MD], BD[MAX_MD], DD[MAX_MD];

int d[MAX_N];

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

    //Bellman-Ford
    for(int k = 0; k < N; k++) {
        for(int i = 0; i + 1 < N; i++) {
            if(d[i + 1] < INF) d[i] = min(d[i], d[i + 1]);
        }

        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]);
        }

        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
        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();
}