poj 2060 Taxi Cab Scheme 最小路径覆盖

//二分匹配的最小路径覆盖

//对于第i次ride，如果在第i次ride结束后还能在第j次ride出发前赶到第j次的出发点

//那么i到j就有一条边

//根据最小路径覆盖 = N - 最大匹配即可得到答案

#include<iostream>

#include<cstdio>

#include<cstring>

#include<cstdlib>

using namespace std;

const int maxn  = 510;

int line[maxn][maxn];

int match[maxn] ;

int vis[maxn] ;

int N;

int find(int start)

{

for(int i = 1 ; i <= N ; i++)

{

if(!vis[i] && line[start][i])

{

vis[i] = 1;

if(match[i] == -1 || find(match[i]))

{

match[i] = start ;

return 1;

}

}

}

return 0;

}

void Match()

{

memset(match , -1 ,sizeof(match)) ;

int ans = 0;

for(int i = 1 ; i <= N ;i++)

{

memset(vis, 0 ,sizeof(vis));

if(find(i))

ans ++ ;

}

printf("%d\n", N-ans);

}

struct node

{

int time_st;

int time_en;

int a,b,c,d;

}ride[maxn];

int main()

{

//freopen("in.txt" , "r" ,stdin);

int T;

scanf("%d", &T);

int h , m ;

while(T--)

{

memset(line, 0 ,sizeof(line));

scanf("%d", &N);

for(int i = 1 ; i <= N ;i++)

{

scanf("%d:%d",&h ,&m);

// cout<<h<<" "<<m<<endl;

ride[i].time_st = 60*h + m;

scanf("%d %d %d %d" , &ride[i].a ,&ride[i].b,&ride[i].c,&ride[i].d);

ride[i].time_en = abs(ride[i].a-ride[i].c) + abs(ride[i].b-ride[i].d) + ride[i].time_st;

for(int j = 1 ; j <= i ;j++)

{

int temp_1 = abs(ride[j].a-ride[i].c) + abs(ride[j].b-ride[i].d);

int temp_2 = abs(ride[i].a-ride[j].c) + abs(ride[i].b-ride[j].d);

if(ride[i].time_en+temp_1 < ride[j].time_st)

line[i][j] = 1;

else if(ride[j].time_en+temp_2 < ride[i].time_st)

line[j][i] = 1;

}

}

Match();

}

return 0;

}