lanqiao 1460 路径(floyd)

12lanqiao/test5-2.cpp

@@ -0,0 +1,35 @@
+// lanqiao 1460 路径(floyd)
+#include<bits/stdc++.h>
+using namespace std;
+const int N = 3e3 + 10, INF = INT_MAX;
+bool vis[N]; // 标记数组
+int mindis[N][N]; // 最短路数组
+int s = 1, n = 2021;
+
+// floyd求多源最短路-时间复杂度-O(n^3)
+void floyd(){
+    // 枚举中转点
+    for(int k = 1; k <= n; k++){
+        // 枚举任意两点
+        for(int i = 1; i <= n; i++){
+            for(int j = 1; j <= n; j++){
+                if(mindis[i][k] != INF && mindis[k][j] != INF){
+                    if(mindis[i][k] + mindis[k][j] < mindis[i][j]){
+                        mindis[i][j] = mindis[i][k] + mindis[k][j];
+                    }
+                }
+            }
+        }
+    }
+}
+int main(){
+    fill(mindis[0], mindis[0] + N * N, INF);
+    for(int i = 1; i <= n; i++){
+        for(int j = max(i-21, 1); j <= min(i+21, 2021); j++){
+            if(i != j) mindis[i][j] = mindis[j][i] = lcm(i, j);
+        }
+    }
+    floyd();
+    cout << mindis[1][n] << endl;
+    return 0;
+}