uvalive 4851 Restaurant（扫描法）

标签：杂题

题意：有一个M*M的网格，坐标[0...M-1,0...M-1] 网格里面有两个y坐标相同的宾馆A和B，以及n个餐厅，宾馆AB里面各有一个餐厅，编号1,2，其他餐厅编号3-n，现在你打算新开一家餐厅，需要考察一下可能的位置，一个位置p是“好位置”的条件是：当且仅当对于已有的每个餐厅q，要么p比q离A近，要么p比q离B近，即dist(p,A) < dist(q,A) 或者 dist(p,B) < dist(q,B)  问“好位置”的个数

#include<cstdio>  
#include<cstring>  
#include<cmath>  
#include<cstdlib>  
#include<iostream>  
#include<algorithm>  
#include<vector>  
#include<map>  
#include<queue>  
#include<stack> 
#include<string>
#include<map> 
#include<set>
#define eps 1e-6 
#define LL long long  
using namespace std;  

const int maxn = 50000 + 50;
const int maxm = 60000 + 50;
const int INF = 0x3f3f3f3f;
int borderl[maxm], borderr[maxm], rest[maxm]; 
int m, n, begx, endx, level;//begx，endx记录ab的横坐标 

void init() {
	memset(rest, -1, sizeof(rest));
	cin >> m >> n;
	int tx, ty;
	cin >> tx >> ty; rest[tx] = ty; begx = tx;
	cin >> tx >> ty; rest[tx] = ty; endx = tx;
	if(begx > endx) swap(begx, endx);
	level = ty;
	for(int i = 3; i <= n; i++) {
		cin >> tx >> ty;
		if(ty < level) ty = 2*level - ty;
		if(rest[tx] != -1) rest[tx] = min(rest[tx], ty);
		else rest[tx] = ty;
	} 
}

void solve() {
	int ans = 0;
	borderl[begx] = borderl[endx] = level;
	borderr[begx] = borderr[endx] = level;
	for(int i = begx + 1; i < endx; i++) {
		if(rest[i] != -1) {
			borderl[i] = min(borderl[i-1]+1, rest[i]);
		} 
		else {
			borderl[i] = borderl[i-1] + 1;
		}
	}
	for(int i = endx - 1; i > begx; i--) {
		if(rest[i] != -1) {
			borderr[i] = min(borderr[i+1]+1, rest[i]);
		} 
		else {
			borderr[i] = borderr[i+1] + 1;
		}
	}
	for(int i = begx + 1; i < endx; i++)
		 ans += max(0, min(min(borderl[i], borderr[i]), m)-max(max(2*level-borderl[i], 2*level-borderr[i]), -1)-1);
	cout << ans << endl;
//	for(int i = begx + 1; i < endx; i++) cout << min(borderl[i], borderr[i]) << " " << rest[i] << endl;
}

int main() {
	//freopen("input.txt", "r", stdin);
	int t; cin >> t;
	while(t--) {
		init();
		solve();
	}
}

uvalive 4851 Restaurant（扫描法）

标签：杂题

原文地址：http://blog.csdn.net/u014664226/article/details/46637319