使用dijkstra算法计算最短路径数和最短路径长度
例1：1003 Emergency：
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.
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.
解：
import java.util.Scanner; public class Main { private static final int MAX_required =505; private static final int MAX_int = Integer.MAX_VALUE; static int[][] map = new int[MAX_required][MAX_required]; static int[] callus = new int[MAX_required]; static class City{ int dist;//起点城市离该城市的距离 boolean visited;//有没有当过pos int number;//起点城市到该城市的最短路径数 int call;//起点城市离该城市的距离 } static City[] city = new City[MAX_required]; public static void main(String[] args) { int n,m,start,end; for (int i = 0; i < MAX_required; i++) { for (int j = 0; j < MAX_required; j++) { if(i == j) { map[i][j] = 0; continue; } map[i][j] = MAX_int; } } Scanner in = new Scanner(System.in); n = in.nextInt(); m = in.nextInt(); start = in.nextInt(); end = in.nextInt(); for (int i = 0; i < n; i++) { callus[i] = in.nextInt(); } for (int i = 0; i < m; i++) { int x = in.nextInt(); int y = in.nextInt(); int num = in.nextInt(); map[x][y] = num; map[y][x] = num; } Dijkstra(start, end, n); } /** * * @param start 开始节点 * @param end 结束节点 * @param n 所有节点数 */ static void Dijkstra(int start, int end, int n) { for (int i = 0; i < n; i++) { city[i] = new City(); city[i].dist = MAX_int; city[i].visited = false; city[i].number = 0; city[i].call = 0; } city[start].dist = 0; city[start].number = 1; city[start].call = callus[start]; for (int cnt = 0; cnt < n; cnt++) { int Min = MAX_int, pos = -1; for (int i = 0; i < n; i++) { if ((city[i].visited == false) && (city[i].dist < Min)) { Min = city[i].dist; pos = i; } } if (pos == -1) { break; } city[pos].visited = true; for (int j = 0; j < n; j++) { if ((city[j].visited == false) && (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; city[j].call = city[pos].call + callus[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 + callus[j]) { city[j].call = city[pos].call + callus[j]; } } } } } System.out.println(city[end].number + " " + city[end].call); } }
图解：
cnt为循环次数，对于连通图中的每个节点，代码都会循环一遍作为中间节点。核心思想为，根据起始节点，cnt循环的每轮确定一个当前已知最短路径的节点。直接看图：
round 1：
当前pos：A
所有pos：A
A到A的最短路径：0
ABCDEpos到各个节点的距离/121N
如果插入pos节点，起点到各个节点的路径0121N
判断后的起点到各个节点的最短路径0121Nround2：
当前pos：B
所有pos：A B
A到B的最短路径：1
ABCDEpos到各个节点的距离//1NN如果插入pos节点，起点到各个节点的路径//2NN
起点到各个节点的最短路径0121Nround3：
当前pos：D
所有pos：A B D
A到D的最短路径：1
ABCDEpos到各个节点的距离//N/1如果插入pos节点，起点到各个节点的路径//N/2
起点到各个节点的最短路径01212round4：
当前pos：C
所有pos：A B D C
A到C的最短路径：2
ABCDEpos到各个节点的距离////1如果插入pos节点，起点到各个节点的路径////3
起点到各个节点的最短路径01212
注：在计算的过程中除了路径还可以加上相同长度路径数、救援人员等参数，在第二重循环中对这些参数进行判断修改，实现最短路径数、最多救援人员等多样性的条件。
如果需要打印出最短路径，如题目《All Roads Lead to Rome》，则可以加个属性，找出当前最短路径（最合适节点）后，记录pos，作为最短路径中该节点的上个节点。
