'commit'

TangDou/AcWing_TiGao/T3/Floyd/344_2.cpp

@@ -0,0 +1,73 @@
+#include <bits/stdc++.h>
+using namespace std;
+const int N = 110;
+#define INF 0x7ffffff
+/**
+分析:模板题，理解floyd 的在 i , j  路径中没有包含k（因为此时k未用来更新），即可写出最小环
+在INF这里wa了两发，习惯值 1e9 不是最大值（查找时判断一下可以避免，但。。。懒了）
+*/
+int n, m;
+int mp[N][N];
+int g[N][N];
+int path[N][N];
+int ans[N];
+int cnt;
+int mm;
+void floyd() {
+    mm = INF;
+    for (int k = 1; k <= n; k++) {
+        for (int i = 1; i < k; i++) {
+            for (int j = i + 1; j < k; j++) {
+                int x = g[i][j] + mp[k][i] + mp[k][j];
+                if (x < mm) {
+                    mm = x;
+                    int tmp = j;
+                    cnt = 0;
+
+                    while (tmp != i) {
+                        ans[cnt++] = tmp;
+                        tmp = path[i][tmp];
+                    }
+                    ans[cnt++] = i;
+                    ans[cnt++] = 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]) {
+                    g[i][j] = g[i][k] + g[k][j];
+                    path[i][j] = path[k][j];
+                }
+            }
+        }
+    }
+}
+
+int main() {
+    while (cin >> n >> m) {
+        for (int i = 1; i <= n; i++) {
+            for (int j = 1; j <= n; j++) {
+                g[i][j] = mp[i][j] = INF;
+                path[i][j] = i;
+            }
+        }
+
+        while (m--) {
+            int a, b, c;
+            cin >> a >> b >> c;
+            g[a][b] = g[b][a] = min(g[a][b], c); // 防重边
+            mp[a][b] = mp[b][a] = g[a][b];
+        }
+
+        floyd();
+
+        if (mm == INF) {
+            puts("No solution.");
+            continue;
+        }
+        for (int i = 0; i < cnt; ++i) printf("%d%s", ans[i], (i == cnt - 1) ? "\n" : " ");
+    }
+    return 0;
+}

TangDou/Topic/【Floyd专题】.md

@@ -180,6 +180,9 @@ int main() {
     return 0;
 }
 ```
+**$TODO$**
+据说可以使用$Dijkstra$算法解决，有空可以试试: **[链接](https://blog.csdn.net/K_R_forever/article/details/80525757)**
+
 
 
 https://juejin.cn/post/6935691567696969764