poj2060

最小路径覆盖，把每个ride看成一个点，如果两个ride可以先后由一辆车完成，那么就从前一个ride引一条边到后一个ride。

在最小路经覆盖中，原图的邻接矩阵和二分图匹配的邻接矩阵为同一矩阵

View Code

#include <iostream>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cmath>

using namespace std;



#define maxn 505



struct Elem

{

    int x1, y1, x2, y2;

    int s, e;

} ride[maxn];



int n;

int uN, vN;

bool g[maxn][maxn];

int xM[maxn], yM[maxn];

bool chk[maxn];



int cal(int h, int m)

{

    return h * 60 + m;

}



int dist(int x1, int y1, int x2, int y2)

{

    return abs(x1 - x2) + abs(y1 - y2);

}



void input()

{

    scanf("%d", &n);

    for (int i = 0; i < n; i++)

    {

        int h, m;

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

        ride[i].s = cal(h, m);

        scanf("%d%d%d%d", &ride[i].x1, &ride[i].y1, &ride[i].x2, &ride[i].y2);

        ride[i].e = ride[i].s + dist(ride[i].x1, ride[i].y1, ride[i].x2, ride[i].y2);

    }

}



bool ok(Elem &a, Elem &b)

{

    return dist(a.x2, a.y2, b.x1, b.y1) + a.e < b.s;

}



void make()

{

    for (int i = 0; i < n; i++)

        for (int j = 0; j < n; j++)

            g[i][j] = ok(ride[i], ride[j]);

}



bool SearchPath(int u)

{

    int v;

    for (v = 0; v < vN; v++)

        if (g[u][v] && !chk[v])

        {

            chk[v] = true;

            if (yM[v] == -1 || SearchPath(yM[v]))

            {

                yM[v] = u;

                xM[u] = v;

                return true;

            }

        }

    return false;

}



int MaxMatch()

{

    int u, ret = 0;

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

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

    for (u = 0; u < uN; u++)

        if (xM[u] == -1)

        {

            memset(chk, false, sizeof(chk));

            if (SearchPath(u))

                ret++;

        }

    return ret;

}



int main()

{

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

    int t;

    scanf("%d", &t);

    while (t--)

    {

        input();

        make();

        uN = vN = n;

        printf("%d\n", n - MaxMatch());

    }

    return 0;

}