'commit'

TangDou/Topic/HDU1385_Precursor.cpp

@@ -1,71 +0,0 @@
-#include <bits/stdc++.h>
-using namespace std;
-
-const int N = 110;
-const int INF = 0x3f3f3f3f;
-// Floyd+记录终点前驱
-int n;
-int g[N][N], w[N];
-int path[N][N]; // 记录i到j最短路径中j的前驱
-
-void floyd() {
-    for (int k = 1; k <= n; k++)
-        for (int i = 1; i <= n; i++)
-            for (int j = 1; j <= n; j++) {
-                if (g[i][j] > g[i][k] + g[k][j] + w[k]) {
-                    g[i][j] = g[i][k] + g[k][j] + w[k];
-                    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
-                }
-                // 相同路径下选择后继更小的(为了字典序)
-                if (g[i][j] == g[i][k] + g[k][j] + w[k])
-                    if (path[i][j] > path[i][k])
-                        path[i][j] = path[i][k];
-            }
-}
-
-// 递归输出路径
-void print(int s, int e) {
-    printf("-->%d", path[s][e]); // 输出s的后继
-    if (path[s][e] != e)         // 如果不是直连
-        print(path[s][e], e);    // 递归输出后继
-}
-
-/*
-From 1 to 3 :
-Path: 1-->5-->4-->3
-Total cost : 21
-
-From 3 to 5 :
-Path: 3-->4-->5
-Total cost : 16
-
-From 2 to 4 :
-Path: 2-->1-->5-->4
-Total cost : 17
-*/
-int main() {
-#ifndef ONLINE_JUDGE
-    freopen("HDU1385.in", "r", stdin);
-#endif
-    while (cin >> n, n) {
-        for (int i = 1; i <= n; i++)
-            for (int j = 1; j <= n; j++) {
-                cin >> g[i][j];
-                if (g[i][j] == -1) g[i][j] = INF;
-                path[i][j] = j;
-            }
-
-        for (int i = 1; i <= n; i++) cin >> w[i];
-        floyd();
-
-        int s, e;
-
-        while (cin >> s >> e, ~s && ~e) {
-            printf("From %d to %d :\n", s, e);
-            printf("Path: %d", s);
-            if (s != e) print(s, e); // 起点与终点不同开始递归
-            printf("\nTotal cost : %d\n\n", g[s][e]);
-        }
-    }
-    return 0;
-}