'commit'

TangDou/AcWing/ErFenTu/POJ3216.cpp

@@ -0,0 +1,72 @@
+#include <iostream>
+#include <algorithm>
+#include <queue>
+#include <map>
+#include <cstring>
+#include <vector>
+#include <stack>
+#include <cstdio>
+using namespace std;
+const int N = 25;
+const int M = 205;
+const int INF = 0x3f3f3f3f;
+struct Task {
+    int id;   // 维修站
+    int cost; // 花费时间
+    int st;   //  起始时间
+} task[M];
+
+int m, n, g[N][N];
+
+// 匈牙利算法
+int match[M], st[M];
+bool dfs(int u) {
+    for (int i = 0; i < m; i++) {
+        if (st[i]) continue;
+        // 这里很妙，不是真的把图建出来，而是直接利用原来的邻接矩阵，通过条件判断来决策，减少了代码!
+        if (task[u].st + task[u].cost + g[task[u].id][task[i].id] > task[i].st) continue;
+
+        st[i] = 1;
+        if (match[i] == -1 || dfs(match[i])) {
+            match[i] = u;
+            return 1;
+        }
+    }
+    return 0;
+}
+
+int main() {
+#ifndef ONLINE_JUDGE
+    freopen("POJ3216.in", "r", stdin);
+#endif
+
+    while (~scanf("%d%d", &n, &m) && n + m) {
+        // 题目中给出的不可达值为-1，因为要求最短路，所以设置为INF
+        for (int i = 1; i <= n; i++)
+            for (int j = 1; j <= n; j++) {
+                scanf("%d", &g[i][j]);
+                if (g[i][j] == -1) g[i][j] = INF;
+            }
+
+        // Floyd求任意两点间最短距离
+        for (int k = 1; k <= n; k++)
+            for (int i = 1; i <= n; i++)
+                for (int j = 1; j <= n; j++)
+                    g[i][j] = min(g[i][j], g[i][k] + g[k][j]);
+
+        // m个任务
+        for (int i = 0; i < m; i++)
+            scanf("%d%d%d", &task[i].id, &task[i].st, &task[i].cost);
+
+        // Hungary
+        memset(match, -1, sizeof match);
+        int res = 0;
+        for (int i = 0; i < m; i++) { // 枚举所个任务
+            memset(st, 0, sizeof st);
+            if (dfs(i)) res++;
+        }
+        // 输出结果
+        printf("%d\n", m - res);
+    }
+    return 0;
+}