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UVA 1665 Islands

时间:2017-07-27 14:26:33      阅读:243      评论:0      收藏:0      [点我收藏+]

标签:put   return   line   mon   一个   star   ack   bit   clear   

题意:输入一个n*m矩阵,每一个格子都有一个正整数,再输入T个整数ti,对于每一个ti,输出大于ti的正整数组成多少个四连快

思路:正着做的话事实上相当于删除连通块,而假设反着做的话就相当于变成添加连通块,把格子都编号然后排序。用并查集


#include<bits/stdc++.h>
using namespace std;
const int maxn = 1005;
const int maxq = 1e5+1;
int pre[maxn*maxn];
int Find(int x)
{
	return x==pre[x]?x:pre[x]=Find(pre[x]);
}
struct Node
{
	int x,y;
	int val;
}nodes[1005*1005];
int mapp[maxn][maxn];
int qq[maxq];
int n,m;
int dir[4][2]={{0,1},{0,-1},{1,0},{-1,0}};
bool cmp(Node a,Node b)
{
	return a.val<b.val;
}
int main()
{
    int T;
	scanf("%d",&T);
	while (T--)
	{
		scanf("%d%d",&n,&m);
		for (int i = 0;i<n;i++)
			for (int j = 0;j<m;j++)
			{
				scanf("%d",&mapp[i][j]);
				int pos = i*m+j;
				nodes[pos].x=i;
				nodes[pos].y=j;
				nodes[pos].val=mapp[i][j];
			}
		int ans = 0;
		int q;
		scanf("%d",&q);
		for (int i = 0;i<q;i++)
			scanf("%d",&qq[i]);
		memset(pre,-1,sizeof(pre));
		sort(nodes,nodes+n*m,cmp);
		int k = n*m-1;
		for (int i = q-1;i>=0;i--)
		{
			if (qq[i]< nodes[k].val)
			{
				while (k>=0 && qq[i]<nodes[k].val)
				{
					int pos = nodes[k].x*m+nodes[k].y;
                    if (!~pre[pos])
						pre[pos]=pos,ans++;
					for (int di=0;di<4;di++)
					{
                        int dx = nodes[k].x+dir[di][0];
						int dy = nodes[k].y+dir[di][1];
						if (dx>=0 && dx<n&&dy>=0&&dy<m&&mapp[dx][dy]>qq[i])
						{
							int ppos = dx*m+dy;
						    if (~pre[ppos])
						    {
							   int u = Find(ppos);
							   int v = Find(pos);
							   if (u!=v)
							    	pre[u]=v,ans--;
						    }
						}
					}
					k--;
				}
				if (k<0)
				{
					for (;i>=0;i--)
					{
						qq[i]=ans;
					}
					break;
				}
		  	}
			qq[i]=ans;
		 }
	    for (int i = 0;i<q;i++)
			printf("%d ",qq[i]);
		printf("\n");
	}
}


Description

技术分享

Deep in the Carribean, there is an island even stranger than the Monkey Island, dwelled by Horatio Torquemada Marley. Not only it has a rectangular shape, but is also divided into an nxm grid. Each grid field has a certain height. Unfortunately, the sea level started to raise and in year i, the level is i meters. Another strange feature of the island is that it is made of sponge, and the water can freely flow through it. Thus, a grid field whose height is at most the current sea level is considered flooded. Adjacent unflooded fields (i.e., sharing common edge) create unflooded areas. Sailors are interested in the number of unflooded areas in a given year.

An example of a x 5 island is given below. Numbers denote the heights of respective fields in meters. Unflooded fields are darker; there are two unflooded areas in the first year and three areas in the second year.

技术分享

Input 

The input contains several test cases. The first line of the input contains a positive integer Z技术分享20, denoting the number of test cases. ThenZ test cases follow, each conforming to the format described below.


The first line contains two numbers n and m separated by a single space, the dimensions of the island, where 1技术分享nm技术分享1000. Next nlines contain m integers from the range [1, 109] separated by single spaces, denoting the heights of the respective fields. Next line contains an integer T(1技术分享T技术分享105). The last line contains T integers tj, separated by single spaces, such that 0技术分享t1技术分享t2技术分享...技术分享tT-1技术分享tT技术分享109.

Output 

For each test case, your program has to write an output conforming to the format described below.


Your program should output a single line consisting of T numbers rj separated by single spaces, where rj is the number of unflooded areas in year tj.

Sample Input 

1 
4 5 
1 2 3 3 1
1 3 2 2 1
2 1 3 4 3
1 2 2 2 2
5 
1 2 3 4 5

Sample Output 

2 3 1 0 0


UVA 1665 Islands

标签:put   return   line   mon   一个   star   ack   bit   clear   

原文地址:http://www.cnblogs.com/cynchanpin/p/7244348.html

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