From 5e156dfb916ba7a417fdac12f4edb986a0f33c26 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=BB=84=E6=B5=B7?= <10402852@qq.com>
Date: Fri, 5 Jan 2024 08:59:01 +0800
Subject: [PATCH] 'commit'

---
 TangDou/Topic/HDU1385.in                      | 12 +++
 TangDou/Topic/HDU1385_Precursor.cpp           | 71 ++++++++++++++++
 .../{HDU1385.cpp => HDU1385_Subsequent.cpp}   | 15 ++++
 TangDou/Topic/【Floyd专题】.md            | 80 ++++++++-----------
 4 files changed, 133 insertions(+), 45 deletions(-)
 create mode 100644 TangDou/Topic/HDU1385.in
 create mode 100644 TangDou/Topic/HDU1385_Precursor.cpp
 rename TangDou/Topic/{HDU1385.cpp => HDU1385_Subsequent.cpp} (85%)

diff --git a/TangDou/Topic/HDU1385.in b/TangDou/Topic/HDU1385.in
new file mode 100644
index 0000000..722ece9
--- /dev/null
+++ b/TangDou/Topic/HDU1385.in
@@ -0,0 +1,12 @@
+5
+0 3 22 -1 4
+3 0 5 -1 -1
+22 5 0 9 20
+-1 -1 9 0 4
+4 -1 20 4 0
+5 17 8 3 1
+1 3
+3 5
+2 4
+-1 -1
+0
\ No newline at end of file
diff --git a/TangDou/Topic/HDU1385_Precursor.cpp b/TangDou/Topic/HDU1385_Precursor.cpp
new file mode 100644
index 0000000..3c4d321
--- /dev/null
+++ b/TangDou/Topic/HDU1385_Precursor.cpp
@@ -0,0 +1,71 @@
+#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;
+}
\ No newline at end of file
diff --git a/TangDou/Topic/HDU1385.cpp b/TangDou/Topic/HDU1385_Subsequent.cpp
similarity index 85%
rename from TangDou/Topic/HDU1385.cpp
rename to TangDou/Topic/HDU1385_Subsequent.cpp
index c04c083..9f519e4 100644
--- a/TangDou/Topic/HDU1385.cpp
+++ b/TangDou/Topic/HDU1385_Subsequent.cpp
@@ -29,8 +29,23 @@ void print(int s, int e) {
     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++) {
diff --git a/TangDou/Topic/【Floyd专题】.md b/TangDou/Topic/【Floyd专题】.md
index c9a9199..10e5f57 100644
--- a/TangDou/Topic/【Floyd专题】.md
+++ b/TangDou/Topic/【Floyd专题】.md
@@ -227,65 +227,55 @@ $i \rightarrow a \rightarrow b \rightarrow c \rightarrow d \rightarrow j$,则$pa
 ```cpp {.line-numbers}
 #include <bits/stdc++.h>
 using namespace std;
+
+const int N = 110;
 const int INF = 0x3f3f3f3f;
+// Floyd+记录起点后继
+int n;
+int g[N][N], w[N];
+int path[N][N]; // 记录i到j最短路径中i的后继
 
-const int N = 1003;
-int g[N][N];    // 邻接矩阵
-int n;          // n个点
-int w[N];       // 额外费用
-int path[N][N]; // i->j 可能存在多条路线，我要找最短的。如果有多条最短的，我要字典序最小的。现在路线唯一了吧！比如这条路线最终是
-// i->a->b->c->d->j,则path[i][j]=a,也就是第一个后继节点。
 void floyd() {
     for (int k = 1; k <= n; k++)
         for (int i = 1; i <= n; i++)
-            if (g[i][k] != INF) // floyd优化
-                for (int j = 1; j <= n; j++) {
-                    if (g[i][j] > g[i][k] + g[k][j] + w[k]) { // w[k]:点权
-                        g[i][j] = g[i][k] + g[k][j] + w[k];   // k的加入，使得i->j的路径变短
-                        path[i][j] = path[i][k];              // 如果i->k->j使得i->j更近，那么根据定义path[i][j]就是这条最短路径中距离i最近的那个点，而这个点由于是出现在i->k的必经之路上，而且是i->k的首席弟子，所以，也必然是i->j的首席弟子。
-                    }
-                    // 处理字典序
-                    if (g[i][j] == g[i][k] + g[k][j] + w[k]) {                // 如果存在多条最短路径，也就是，除了k还有其它k1,k2使得i->j距离一样小
-                        if (path[i][j] > path[i][k]) path[i][j] = path[i][k]; // 字典序，谁更小就留下谁
-                    }
+            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
 }
+
 int main() {
-    while (cin >> n && n) {
-        for (int i = 1; i <= n; i++) {
+    while (cin >> n, n) {
+        for (int i = 1; i <= n; i++)
             for (int j = 1; j <= n; j++) {
-                path[i][j] = j;                   // 路径初始化,记录整条路径上，离i节点最近的，最短路径上的下一个点，只有i->j时，下一个点可不就是j
-                cin >> g[i][j];                   // 不管是不是有边，都先录进来
-                if (g[i][j] == -1) g[i][j] = INF; // 如果题目中给出的是无边，那么设置为正无穷。此时，有些记录的path[i][j]就是没用的，但没事，后面会被其它代码替换掉path[i][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]; // 读入点权
 
-        // 多源最短路径
+        for (int i = 1; i <= n; i++) cin >> w[i];
         floyd();
 
-        // 处理询问
-        int x, y;
-        while (cin >> x >> y) {
-            if (x == -1 && y == -1) break;
-            printf("From %d to %d :\n", x, y);
-            printf("Path: %d", x);
-            int u = x, v = y;
-            // 理解路径思路:
-            // (1) 从起点x出发,用循环打印路径,最后一个打印的肯定是y
-            // (2) 从起点x出发,第二个点应该是离x最近的，并且是最短路径上的那个点,这个点就是path[x][y]!
-            // path[x][y]：从起点x出发，到终点y有多条最短路径，我们选择字典序最小的那条最短路径，然后path[x][y]就是从x出发，离x最近的这条最短路径上的点。
-            while (x != y) {
-                printf("-->%d", path[x][y]); // 输出距离x最近的那个点
-                x = path[x][y];              // 更换x概念，向y逼近，让循环跑起来
-            }
+        int s, e;
 
-            puts("");
-            if (g[u][v] < INF)
-                printf("Total cost : %d\n", g[u][v]);
-            else
-                puts("-1");
-            puts("");
+        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;