Poj-2060 Taxi Cab Scheme 二分图最小路径覆盖

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题目链接

出租车公司有n个预约, 每个预约有时间和地点, 地点分布在二维整数坐标系上, 地点之间的行驶时间为两点间的曼哈顿距离（|x1 - x2| + |y1 - y2|）。一辆车可以在运完一个乘客后运另一个乘客, 条件是此车要在预约开始前一分钟之前到达出发地, 问最少需要几辆车搞定所有预约。

#include <stdio.h>#include <string.h>#include <math.h>#include <iostream>#include <algorithm>using namespace std;typedef long long LL;const int maxn = 505;const int Mod = 1000000007;const double inf = 1<<30;int n;int map[maxn][maxn];struct node{int sta,t;int sx,sy,ex,ey;}book[maxn];int cx[maxn],cy[maxn];bool vis[maxn];bool FindPath( int u ){for( int i = u+1; i <= n; i ++ ){if( !vis[i] && map[u][i] ){vis[i] = true;if( cy[i] == -1 || FindPath( cy[i] ) ){cy[i] = u;cx[u] = i;return true;}}}return false;}int MaxMatch(){int ans = 0;memset( cx,-1,sizeof(cx) );memset( cy,-1,sizeof(cy) );for( int i = 1; i <= n; i++ ){if( cx[i] == -1 ){memset( vis,0,sizeof(vis) );ans += FindPath( i );}}return ans;}int main(){#ifndef ONLINE_JUDGE   freopen("data.txt","r",stdin);   #endifint cas,h,m;scanf("%d",&cas);while( cas -- ){scanf("%d",&n);memset( map,0,sizeof(map) );for( int i = 1; i <= n; i ++ ){scanf("%d:%d%d%d%d%d",&h,&m,&book[i].sx,&book[i].sy,&book[i].ex,&book[i].ey);book[i].sta = h*60 + m;book[i].t = abs( book[i].sx - book[i].ex ) + abs( book[i].sy - book[i].ey );}for( int i = 1; i <= n; i ++ ){for( int j = i+1; j <= n; j ++ ){int dis = abs( book[j].sx - book[i].ex ) + abs( book[j].sy - book[i].ey );if( book[i].t + dis + 1 <= book[j].sta - book[i].sta )map[i][j] = 1;}}printf("%d\n",n - MaxMatch());}return 0;}

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