70 lines
1.9 KiB
C++
70 lines
1.9 KiB
C++
#include <bits/stdc++.h>
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using namespace std;
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const int N = 110;
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const int INF = 0x3f3f3f3f;
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// Floyd+记录起点后继
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int n;
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int g[N][N], w[N];
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int path[N][N]; // 记录i到j最短路径中i的后继
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void floyd() {
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for (int k = 1; k <= n; k++)
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for (int i = 1; i <= n; i++)
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for (int j = 1; j <= n; j++) {
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if (g[i][j] > g[i][k] + g[k][j] + w[k]) {
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g[i][j] = g[i][k] + g[k][j] + w[k];
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path[i][j] = path[i][k]; // i->j这条最短路径上，i后面第一个节点，是i->k路径上第一个节点
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}
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// 相同路径下选择后继更小的(为了字典序)
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if (g[i][j] == g[i][k] + g[k][j] + w[k])
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if (path[i][j] > path[i][k])
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path[i][j] = path[i][k];
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}
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}
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// 递归输出路径
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void print(int s, int e) {
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printf("-->%d", path[s][e]); // 输出s的后继
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if (path[s][e] != e) // 如果不是直连
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print(path[s][e], e); // 递归输出后继
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}
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/*
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From 1 to 3 :
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Path: 1-->5-->4-->3
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Total cost : 21
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From 3 to 5 :
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Path: 3-->4-->5
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Total cost : 16
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From 2 to 4 :
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Path: 2-->1-->5-->4
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Total cost : 17
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*/
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int main() {
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#ifndef ONLINE_JUDGE
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freopen("HDU1385.in", "r", stdin);
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#endif
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while (cin >> n, n) {
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for (int i = 1; i <= n; i++)
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for (int j = 1; j <= n; j++) {
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cin >> g[i][j];
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if (g[i][j] == -1) g[i][j] = INF;
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path[i][j] = j;
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}
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for (int i = 1; i <= n; i++) cin >> w[i];
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floyd();
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int s, e;
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while (cin >> s >> e, ~s && ~e) {
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printf("From %d to %d :\n", s, e);
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printf("Path: %d", s);
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if (s != e) print(s, e); // 起点与终点不同开始递归
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printf("\nTotal cost : %d\n\n", g[s][e]);
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}
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}
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return 0;
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} |