UVA 1201 - Taxi Cab Scheme（二分图匹配+最小路径覆盖)

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UVA 1201 - Taxi Cab Scheme

题目链接

题意：给定一些乘客，每个乘客需要一个出租车，有一个起始时刻，起点，终点，行走路程为曼哈顿距离，每辆出租车必须在乘客一分钟之前到达，问最少需要几辆出租车

思路：如果一辆车载完一个乘客a，能去载乘客b，就连一条有向边，这样做完整个图形成一个DAG，然后要求的最少数量就是最小路径覆盖，利用二分图最大匹配去做，把每个点拆成两点，如果有边就连边，做一次最大匹配，n - 最大匹配数就是答案

代码：

#include <cstdio>#include <cstring>#include <algorithm>#include <cstdlib>#include <vector>using namespace std;const int N = 505;int t, n;struct People {int s, x1, y1, x2, y2;void read() {int h, m;scanf("%d:%d%d%d%d%d", &h, &m, &x1, &y1, &x2, &y2);s = h * 60 + m;}bool operator < (const People& c) const {return s < c.s;}} p[N];vector<int> g[N];bool judge(People a, People b) {int tmp = a.s + abs(a.x2 - a.x1) + abs(a.y2 - a.y1) + abs(a.x2 - b.x1) + abs(a.y2 - b.y1);if (tmp < b.s) return true;return false;}int match[N], vis[N];bool dfs(int u) {for (int i = 0; i < g[u].size(); i++) {int v = g[u][i];if (vis[v]) continue;vis[v] = 1;if (match[v] == -1 || dfs(match[v])) {match[v] = u;return true;}}return false;}int hungary() {int ans = 0;memset(match, -1, sizeof(match));for (int i = 0; i < n; i++) {memset(vis, 0, sizeof(vis));if (dfs(i)) ans++;}return ans;}int main() {scanf("%d", &t);while (t--) {scanf("%d", &n);for (int i = 0; i < n; i++) {g[i].clear();p[i].read();}sort(p, p + n);for (int i = 0; i < n; i++)for (int j = i + 1; j < n; j++) {if (judge(p[i], p[j]))g[i].push_back(j);}printf("%d\n", n - hungary());}return 0;}

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