1003. Emergency (25)

分析：

　　考察最短路，可以使用Dijkstra算法，关键在于有相等路径时的判断。

　　特别注意：题目要求的是the number of different shortest paths 和 the maximum amount of rescue teams you can possibly gather。

#include <stdio.h>
#include <iostream>
#include <string>
#include <cctype>
//#include <map>

using namespace std;

const int Max_required = 505;
const int Max_int = 0x7fffffff;

int map[Max_required][Max_required];
int number_of_team_in_city[Max_required];

struct CITY
{
    int dist;
    bool visit;
    int call;
    int number;
}city[Max_required];

void Dijkstra(int start, int end, int n)
{
    for (int i = 0; i < n; i++)
    {
        city[i].dist = Max_int;
        city[i].visit = 0;
        city[i].number = 0;
        city[i].call = 0;
    }
    city[start].dist = 0;
    city[start].call = number_of_team_in_city[start];
    city[start].number = 1;

    //work......
    for (int i = 0; i < n; i++)
    {
        int Min = Max_int, pos = -1;
        for (int j = 0; j < n; j++)
        {
            if (city[j].visit == 0 && city[j].dist < Min)
            {
                Min = city[j].dist;
                pos = j;
            }
        }
        if (pos == -1) break;
        city[pos].visit = 1;

        for (int j = 0; j < n; j++)
        {
            if (city[j].visit == 0 && map[pos][j] != Max_int) 
            {
                if (city[j].dist > city[pos].dist + map[pos][j])
                {
                    city[j].dist = city[pos].dist + map[pos][j];
                    //-------------------------------
                    city[j].number = city[pos].number;    //注意！
                    //---------------------------------
                    city[j].call = city[pos].call + number_of_team_in_city[j];
                }
                else if (city[j].dist == city[pos].dist + map[pos][j])
                {
                    city[j].number += city[pos].number;
                    if (city[j].call < city[pos].call + number_of_team_in_city[j])
                        city[j].call = city[pos].call + number_of_team_in_city[j];
                    //city[j].call += city[pos].call;
                }
            }
        }
    }

    printf("%d %d\n", city[end].number, city[end].call);
    //printf("%d\n", city[end].dist);
}

int main()
{
    int n, m, start, end;

    while (scanf("%d%d%d%d", &n, &m, &start, &end) != EOF)
    {
        for (int i = 0; i < n; i++)
        {
            scanf("%d", &number_of_team_in_city[i]);
            for (int j = 0; j < n; j++)
                map[i][j] = Max_int;
        }

        while (m--)
        {
            int from, to, road_len;
            scanf("%d%d%d", &from, &to, &road_len);

            map[from][to] = road_len;
            map[to][from] = road_len;
        }

        Dijkstra(start, end, n);
    }
    return 0;
}