HDU 5652 India and China Origins（并查集）

India and China Origins

Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 441    Accepted Submission(s): 133

Problem Description

A long time ago there are no himalayas between India and China, the both cultures are frequently exchanged and are kept in sync at that time, but eventually himalayas rise up. With that at first the communation started to reduce and eventually died.

Let's assume from my crude drawing that the only way to reaching from India to China or viceversa is through that grid, blue portion is the ocean and people haven't yet invented the ship. and the yellow portion is desert and has ghosts roaming around so people can't travel that way. and the black portions are the location which have mountains and white portions are plateau which are suitable for travelling. moutains are very big to get to the top, height of these mountains is infinite. So if there is mountain between two white portions you can't travel by climbing the mountain. And at each step people can go to 4 adjacent positions. Our archeologists have taken sample of each mountain and estimated at which point they rise up at that place. So given the times at which each mountains rised up you have to tell at which time the communication between India and China got completely cut off.

Input

There are multi test cases. the first line is a sinle integer  which represents the number of test cases. For each test case, the first line contains two space seperated integers . next  lines consists of strings composed of  characters.  denoting that there's already a mountain at that place,  denoting the plateau. on  line there will be an integer  denoting the number of mountains that rised up in the order of times. Next  lines contain  space seperated integers  denoting that at ith year a mountain rised up at location .

Output

Single line at which year the communication got cut off. print -1 if these two countries still connected in the end. Hint:

From the picture above, we can see that China and India have no communication since 4th year.

Sample Input

1
4 6
011010
000010
100001
001000
7
0 3
1 5
1 3
0 0
1 2
2 4
2 1

Sample Output

4

Source

BestCoder Round #77 (div.2)

用二分加验证可以过，用并查集也可以过。

这个是并查集，

</pre><p style="height:auto; margin:0px; padding:0px 20px; font-size:14px; font-family:'Times New Roman'"><pre name="code" class="html">#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>

using namespace std;
#define MAX 250000
int father[MAX+5];
int dir[4][2]={{0,1},{0,-1},{1,0},{-1,0}};
int find(int x)
{
	if(father[x]!=x)
		father[x]=find(father[x]);
	return father[x];
}
int b[MAX][2];
char a[505][505];
int c[505][505];
int q;
int n,m;

int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d",&n,&m);
		for(int i=1;i<=n;i++)
			scanf("%s",a[i]+1);
		scanf("%d",&q);
		for(int i=1;i<=q;i++)
		{
			scanf("%d%d",&b[i][0],&b[i][1]);
			b[i][0]++;b[i][1]++;
			a[b[i][0]][b[i][1]]='1';
		}
		for(int i=0;i<=n*m+1;i++)
			father[i]=i;
		for(int i=1;i<=m;i++)
		{
			if(a[1][i]=='0') father[i]=0;
			if(a[n][i]=='0') father[(n-1)*m+i]=n*m+1;
		}
		for(int i=2;i<=n;i++)
		{
			for(int j=1;j<=m;j++)
			{
				if(a[i][j]=='1')
					continue;
				if(a[i][j-1]=='0'&&j!=1)
				{
					int fx=find((i-1)*m+j);
					int fy=find((i-1)*m+j-1);
					if(fx!=fy)
						father[fx]=fy;
				}
				if(a[i-1][j]=='0')
				{
					int fx=find((i-1)*m+j);
					int fy=find((i-2)*m+j);
					if(fx!=fy)
						father[fx]=fy;
				}
			}
		}
		
		if(find(0)==find(n*m-1))
		{
			printf("-1\n");
			continue;
		}
		int i;
		for( i=q;i>=1;i--)
		{
			for(int j=0;j<4;j++)
			{
				int x=b[i][0],y=b[i][1];
				if(x==1)
					father[(x-1)*m+y]=0;
				if(x==n)
					father[(x-1)*m+y]=n*m+1;
				int xx=x+dir[j][0];int yy=y+dir[j][1];
				if(xx<1||xx>n||yy<1||yy>m||a[xx][yy]=='1')
					continue;
				int fx=find((x-1)*m+y);
				int fy=find((xx-1)*m+yy);
				if(fx!=fy)
					father[fx]=fy;
			}
			a[b[i][0]][b[i][1]]='0';
			if(find(0)==find(n*m+1))
				break;
		}
		printf("%d\n",i);
	}
	return 0;
}