【最小路徑覆蓋】POJ 2060

題意是求最小用幾輛出租車完成n個任務，方法：最小路徑覆蓋
= 點數 - 最大匹配數，這裡有個注意的細節http://www.cnblogs.com/ka200812/archive/2011/07/31/2122641.html

#include <map>#include <set>#include <list>#include <queue>#include <deque>#include <stack>#include <string>#include <cstdio>#include <math.h>#include <iomanip>#include <cstdlib>#include <limits.h>#include <string.h>#include <iostream>#include <fstream>#include <algorithm>using namespace std;#define LL long long#define MIN -99999999#define MAX 99999999#define pii pair<int ,int>#define bug cout<<"here!!"<<endl#define PI acos(-1.0)#define FRE freopen("input.txt","r",stdin)#define FF freopen("output.txt","w",stdout)#define eps 1e-8#define N 510struct node{    int beg,end;    int x1,y1,x2,y2;}p[N];int match[N];vector<int>  v[N];bool vis[N];bool sear(int s){    int i,j;    for(i=0;i<v[s].size();i++){        int x = v[s][i];        if(!vis[x]){            vis[x] = 1;            if(match[x] == -1 || sear(match[x])){                match[x] = s;                return true;            }        }    }    return false;}int main(){    int t;    scanf("%d",&t);    while(t--){        int n;        scanf("%d",&n);        int i,j;        for(i=0;i<n;i++){            int aa,bb;            scanf("%d:%d %d %d %d %d",&aa,&bb,&p[i].x1,&p[i].y1,&p[i].x2,&p[i].y2);            p[i].beg = aa*60+bb;            p[i].end = p[i].beg+abs(p[i].x1-p[i].x2)+abs(p[i].y1-p[i].y2);            v[i].clear();        }        memset(match,-1,sizeof(match));        for(i=0;i<n;i++){            for(j=0;j<n;j++){                if(i==j)continue;                int dis = abs(p[i].x2 - p[j].x1)+abs(p[i].y2 - p[j].y1);                if(p[i].end+dis<p[j].beg){                    v[i].push_back(j);                }            }        }        int cnt = 0;        for(i=0;i<n;i++){            memset(vis,0,sizeof(vis));            if(sear(i))cnt++;        }        printf("%d\n",n-cnt);    }    return 0;}