Emergency (25分)

最新推荐文章于 2022-01-23 15:54:31 发布

Amber0130

最新推荐文章于 2022-01-23 15:54:31 发布

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分类专栏： Pat

本文链接：https://blog.youkuaiyun.com/weixin_43601103/article/details/112862011

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该博客探讨了一种紧急救援团队领导者如何利用特殊地图快速规划从当前城市到目标城市的最短路径，并在途中最大限度地召集救援队伍。输入包括城市数量、道路数量、起点和终点城市，以及各城市的救援队伍数量和道路长度。输出是最短路径的数量和能聚集的最大救援队伍数。算法基于Dijkstra's算法进行路径搜索，同时考虑了路径条数和资源调度的双重指标。

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As an emergency rescue team leader of a city, you are given a special map of your country. The map shows several scattered cities connected by some roads. Amount of rescue teams in each city and the length of each road between any pair of cities are marked on the map. When there is an emergency call to you from some other city, your job is to lead your men to the place as quickly as possible, and at the mean time, call up as many hands on the way as possible.

Input Specification:

Each input file contains one test case. For each test case, the first line contains $4$ positive integers: $N (\leq 500)$ - the number of cities (and the cities are numbered from $0$ to $N - 1$ ), $M$ - the number of roads, $C_1$ and $C_2$ - the cities that you are currently in and that you must save, respectively. The next line contains $N$ integers, where the $i - t h$ integer is the number of rescue teams in the $i - t h$ city. Then $M$ lines follow, each describes a road with three integers $c_1$ , $c_2$ and $L$ , which are the pair of cities connected by a road and the length of that road, respectively. It is guaranteed that there exists at least one path from $C 1$ to $C 2$

Output Specification:

For each test case, print in one line two numbers: the number of different shortest paths between $C_1$ and $C_2$ , and the maximum amount of rescue teams you can possibly gather. All the numbers in a line must be separated by exactly one space, and there is no extra space allowed at the end of a line.

Sample Input:

Sample Output:

2 4

思路

此题是求最短路径条数加上第二标尺点权的综合应用。
当 $d [u] + d i s = = d [v]$ 时，无论 $m x [u] + a [v] > m x [v]$ 是否成立，都应当让 $n u m [v] + = n u m [u]$ ,因为最短路径条数的依据仅仅是第一标尺距离，与点权无关。

#include <bits/stdc++.h>

using namespace std;
const int maxn = 510;
const int inf = 0x3f3f3f3f;
int n,m,s,t;
struct edge{
    int v,dis;
};
vector<edge>e[maxn];
int a[maxn];
int d[maxn];
bool vis[maxn];
int num[maxn];
int mx[maxn];

void Dijstra(int s) {
    memset(d, 0x3f, sizeof(d));
    d[s] = 0;
    num[s] = 1;
    mx[s]=a[s];
    for (int i = 0; i < n; i++) {
        int u = -1, mi = inf;
        for (int j = 0; j < n; j++) {
            if (!vis[j] && d[j] < mi) {
                mi = d[j];
                u = j;
            }
        }
        if (u == -1) break;
        vis[u] = 1;
        for (auto &[v, dis]:e[u]) {
            if (vis[v]) continue;
            if (d[u] + dis < d[v]) {
                d[v] = d[u] + dis;
                num[v] = num[u];
                mx[v] = mx[u] + a[v];
            } else {
                if (d[u] + dis == d[v]) {
                    num[v] += num[u];
                    mx[v] = max(mx[u] + a[v], mx[v]);
                }
            }
        }
    }
    printf("%d %d\n", num[t], mx[t]);
}
int main() {
    scanf("%d%d%d%d", &n, &m, &s, &t);
    for (int i = 0; i < n; i++) {
        scanf("%d", &a[i]);
    }
    for (int i = 1; i <= m; i++) {
        int u, v, w;
        scanf("%d%d%d", &u, &v, &w);
        e[u].push_back({v, w});
        e[v].push_back({u, w});
    }
    Dijstra(s);
    return 0;
}