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  • POJ2060 Taxi Cab Scheme 出租车(NWERC2004)

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    题面:有(m)位客人从城市的不同位置出发,到达各自目的地。已知每人的出发时间、地点和目的地,用尽量少的出租车送他们,使得每次出租车接客人时,至少能提前一分钟到达他所在的位置。注意,为了满足这一条件,要么这位客人是这辆出租车接送的第一个人,要么在接送完上一个客人后,有足够的时间从上一个目的地开到这里。为简单起见,假定城区是网格型的,地址用坐标((x,y))表示。出租车从((x_1, y_1))处到((x_2,y_2))处需要行驶(|x_1 -x_2| + |y_1 −y_2|)分钟。

    #include<cstdio>
    #include<iostream>
    #include<cmath>
    #include<algorithm>
    #include<cstring>
    #include<cstdlib>
    #include<cctype>
    #include<vector>
    #include<queue>
    #include<assert.h>
    #include<ctime>
    using namespace std;
    #define enter puts("") 
    #define space putchar(' ')
    #define Mem(a, x) memset(a, x, sizeof(a))
    #define In inline
    #define forE(i, x, y) for(int i = head[x], y; ~i && (y = e[i].to); i = e[i].nxt)
    typedef long long ll;
    typedef double db;
    const int INF = 0x3f3f3f3f;
    const db eps = 1e-8;
    const int maxn = 505;
    const int maxe = 3e5 + 5;
    In ll read()
    {
    	ll ans = 0;
    	char ch = getchar(), las = ' ';
    	while(!isdigit(ch)) las = ch, ch = getchar();
    	while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
    	if(las == '-') ans = -ans;
    	return ans;
    }
    In void write(ll x)
    {
    	if(x < 0) x = -x, putchar('-');
    	if(x >= 10) write(x / 10);
    	putchar(x % 10 + '0');
    }
    
    char s[6];
    int n;
    struct Edge
    {
    	int nxt, to;
    }e[maxe];
    int head[maxn], ecnt = -1;
    In void addEdge(int x, int y)
    {
    	e[++ecnt] = (Edge){head[x], y};
    	head[x] = ecnt;
    }
    struct Node
    {
    	int tim, tim2, xs, ys, xe, ye;
    	In void calc(char* s)
    	{
    		tim = ((s[0] - '0') * 10 + (s[1] - '0')) * 60;
    		tim += (s[3] - '0') * 10 + (s[4] - '0');
    		tim2 = abs(xs - xe) + abs(ys - ye);
    	}
    }t[maxn];
    In bool check(int i, int j)
    {
    	return t[i].tim2 + abs(t[i].xe - t[j].xs) + abs(t[i].ye - t[j].ys) < t[j].tim - t[i].tim;
    }
    
    bool vis[maxn];
    int lft[maxn];
    In bool dfs(int now)
    {
    	forE(i, now, v) if(!vis[v])
    	{
    		vis[v] = 1;
    		if(!lft[v] || dfs(lft[v])) {lft[v] = now; return 1;}
    	}
    	return 0;
    }
    In int hung()
    {
    	int ret = 0; Mem(lft, 0);
    	for(int i = 1; i <= n; ++i)
    	{
    		Mem(vis, 0);
    		if(dfs(i)) ++ret;
    	}
    	return ret;
    }
    
    int main()
    {
    	int T = read();
    	while(T--)
    	{
    		Mem(head, -1), ecnt = -1;
    		n = read();
    		for(int i = 1; i <= n; ++i) 
    		{
    			scanf("%s", s);
    			t[i].xs = read(), t[i].ys = read(), t[i].xe = read(), t[i].ye = read();
    			t[i].calc(s);
    		}
    		for(int i = 1; i <= n; ++i)
    			for(int j = i + 1; j <= n; ++j) 
    				if(check(i, j)) addEdge(i, j);	
    		write(n - hung()), enter;
    	}
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/mrclr/p/14775634.html
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