1003. Emergency (25)

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本文介绍了一种用于紧急情况下的救援路径规划算法。该算法能在限定时间内找到从当前城市到目标城市的最短路径，并尽可能多地集结救援队伍。输入包括城市数量、道路连接及救援队伍分布等信息。

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1003. Emergency (25)
时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B
判题程序 Standard 作者 CHEN, Yue

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
Each input file contains one test case. For each test case, the first line contains 4 positive integers: N (<= 500) - the number of cities (and the cities are numbered from 0 to N-1), M - the number of roads, C1 and C2 - the cities that you are currently in and that you must save, respectively. The next line contains N integers, where the i-th integer is the number of rescue teams in the i-th city. Then M lines follow, each describes a road with three integers c1, c2 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 C1 to C2.
Output
For each test case, print in one line two numbers: the number of different shortest paths between C1 and C2, 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
5 6 0 2
1 2 1 5 3
0 1 1
0 2 2
0 3 1
1 2 1
2 4 1
3 4 1

#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
#include <vector>
#include <queue>
#include <cfloat>
#include <map>

#define INF INT32_MAX

using namespace std;
const int MaxN = 510;
int rescue[MaxN],w[MaxN];
int roadNum[MaxN];
int des[MaxN], G[MaxN][MaxN];
bool visited[MaxN] = { false };
int N;


void Dijkstra(int s)
{
    fill(roadNum, roadNum + N, 0);
    fill(des, des + N, INF);
    fill(w, w + N, INF);
    des[s] = 0; roadNum[s] = 1;
    w[s] = rescue[s];

    for (int i = 0,Min; i < N; ++i)
    {
        Min = INF; int u = -1;
        for (int k = 0; k < N; ++k)
        {
            if (!visited[k] && Min > des[k])
            {
                Min = des[k];
                u = k;
            }
        }

        if (u == -1)return;
        visited[u] = true;

        for (int v = 0; v < N; ++v)
        {
            if (!visited[v] && G[u][v] != INF)
            {
                if (des[v] > G[u][v] + des[u])
                {
                    des[v] = G[u][v] + des[u];
                    roadNum[v] = roadNum[u];

                    w[v] = rescue[v] + w[u];
                }
                else if (des[v] == G[u][v] + des[u])
                {
                    roadNum[v] += roadNum[u];

                    if (w[v] < rescue[v] + w[u])
                        w[v] = rescue[v] + w[u];
                }
            }
        }
    }
}

int main()
{
#ifdef _DEBUG
    freopen("data.txt", "r+", stdin);
#endif // _DEBUG

    std::ios::sync_with_stdio(false);

    int L, s, e;
    cin >> N >> L >> s >> e;
    fill(G[0], G[0] + (N + 1) * MaxN, INF);

    for (int i = 0; i < N; ++i)
        cin >> rescue[i];

    for (int k = 0; k < L; ++k)
    {
        int u, v, w;
        cin >> u >> v >> w;
        G[u][v] = G[v][u] =  w;
    }

    Dijkstra(s);

    cout << roadNum[e] << " " << w[e];

    return 0;
}