'commit'

TangDou/Topic/P1828.cpp

@@ -0,0 +1,38 @@
+#include <bits/stdc++.h>
+using namespace std;
+const int N = 814;
+const int INF = 0x3f3f3f3f;
+int id[N];
+int a, b, c, g[N][N], res = INF;
+int main() {
+    // 加快读入
+    ios::sync_with_stdio(false), cin.tie(0);
+    int p, n, m; // p只奶牛，n个牧场，m条边
+    cin >> p >> n >> m;
+
+    memset(g, 0x3f, sizeof g);
+    for (int i = 1; i <= n; i++) g[i][i] = 0; // 初始化
+
+    for (int i = 1; i <= p; i++) cin >> id[i]; // i号奶牛，在id[i]这个牧场
+
+    while (m--) {
+        cin >> a >> b >> c;
+        g[a][b] = g[b][a] = min(c, g[a][b]);
+    }
+    // 标准的Floyd板子
+    for (int k = 1; k <= n; k++)
+        for (int i = 1; i <= n; i++) {
+            if (g[i][k] == INF) continue; // floyd小优化
+            for (int j = 1; j <= n; j++)
+                if (g[i][j] > g[i][k] + g[k][j])
+                    g[j][i] = g[i][j] = g[i][k] + g[k][j];
+        }
+
+    for (int i = 1; i <= n; i++) { // 每个牧场出发
+        int ans = 0;
+        for (int j = 1; j <= p; j++) ans += g[i][id[j]];
+        res = min(res, ans);
+    }
+    printf("%d", res);
+    return 0;
+}

TangDou/Topic/SSL_1760.cpp

@@ -0,0 +1,29 @@
+#include <bits/stdc++.h>
+using namespace std;
+const int N = 210;
+int n, g[N][N];
+const int INF = 0x3f3f3f3f;
+int ans = INF;
+int main() {
+    cin >> n;
+    for (int i = 1; i <= n; i++)
+        for (int j = 1; j <= n; j++) {
+            cin >> g[i][j];
+            if (g[i][j] == 0 && i != j) g[i][j] = INF; // 建图(注意i==j要为0)
+        }
+    // 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])
+                    g[i][j] = g[i][k] + g[k][j];
+
+    int s;
+    for (int i = 1; i <= n; i++) {
+        s = 0;
+        for (int j = 1; j <= n; j++) s += g[i][j];
+        if (s < ans) ans = s;
+    }
+    printf("%d", ans);
+    return 0;
+}

TangDou/Topic/【Floyd专题】.md

@@ -378,4 +378,77 @@ int main() {
     }
     return 0;
 }
-```
+```
+
+### 八、其它习题
+
+#### $SSL-1760$（商店选址）
+**题目**
+给出一个城市的地图（用邻接矩阵表示），商店设在一点，使各个地方到商店距离之和最短。
+
+$Input$
+第一行为$n$（共有几个城市）； $N$小于$201$
+第二行至第$n+1$行为城市地图（用邻接矩阵表示）
+
+$Output$
+最短路径之和
+
+$Sample$ $Input$
+```cpp {.line-numbers}
+3
+0 3 1
+3 0 2
+1 2 0
+1
+2
+3
+4
+```
+
+$Sample$ $Output$
+```cpp {.line-numbers}
+3
+1
+```
+```cpp {.line-numbers}
+#include <bits/stdc++.h>
+using namespace std;
+const int N = 210;
+int n, g[N][N];
+const int INF = 0x3f3f3f3f;
+int ans = INF;
+int main() {
+    cin >> n;
+    for (int i = 1; i <= n; i++)
+        for (int j = 1; j <= n; j++) {
+            cin >> g[i][j];
+            if (g[i][j] == 0 && i != j) g[i][j] = INF; // 建图(注意i==j要为0)
+        }
+    // 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])
+                    g[i][j] = g[i][k] + g[k][j];
+
+    int s;
+    for (int i = 1; i <= n; i++) {
+        s = 0;
+        for (int j = 1; j <= n; j++) s += g[i][j];
+        if (s < ans) ans = s;
+    }
+    printf("%d", ans);
+    return 0;
+}
+
+```
+#### [$P1828$ [$USACO3.2$] 香甜的黄油 $Sweet$ $Butter$](https://www.luogu.com.cn/problem/P1828)
+
+
+SSL-1761（城市问题）
+
+#### [$P1364$ 医院设置](https://www.luogu.com.cn/problem/P1364)
+
+SSL-1613=CCF1342（最短路径问题）
+
+