HDU 5652 India and China Origins 二分+並查集


India and China Origins

題目連接:

http://acm.hdu.edu.cn/showproblem.php?pid=5652

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 T which represents the number of test cases.

For each test case, the first line contains two space seperated integers N,M. next N lines consists of strings composed of 0,1 characters. 1 denoting that there's already a mountain at that place, 0 denoting the plateau. on N+2 line there will be an integer Q denoting the number of mountains that rised up in the order of times. Next Q lines contain 2 space seperated integers X,Y denoting that at ith year a mountain rised up at location X,Y.

T≤10

1≤N≤500

1≤M≤500

1≤Q≤N∗M

0≤X<N

0≤Y<M

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

Hint

題意

給一個n*m的圖

然后1表示有障礙物,中國在最上面,印度在最下面

然后有q個變化,每次變化都會有一個地方從0變成障礙物

問你最早什么時候,中國和印度不相連了。

題解:

比較顯然的就是二分+判斷是否連通

判斷是否連通用並查集和bfs都可以

反正都是O(n)嘛

代碼

#include<stdio.h>
#include<iostream>
#include<math.h>
#include<cstring>
using namespace std;
const int maxn = 505;
int p[505*505],n,m,q;
char s[maxn][maxn];
int vis[maxn][maxn];
pair<int,int>d[maxn*maxn];
int dx[4]={1,-1,0,0};
int dy[4]={0,0,1,-1};
int fi(int x)
{
    return p[x]==x?x:p[x]=fi(p[x]);
}
void uni(int p1,int q1)
{
    int x=fi(p1);
    int y=fi(q1);
    if(x!=y)
        p[x]=y;
}
int check(int t)
{
    for(int i=0;i<=n*m+1;i++)p[i]=i;
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
            if(s[i][j]=='1')vis[i][j]=1;
            else vis[i][j]=0;
    for(int i=1;i<=t;i++)
        vis[d[i].first][d[i].second]=1;
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            for(int k=0;k<4;k++)
            {
                if(vis[i][j]==1)continue;
                int xx = i+dx[k];
                int yy = j+dy[k];
                if(yy<=0||yy>m)continue;
                if(xx==0)uni((i-1)*m+j,0);
                else if(xx==n+1)uni((i-1)*m+j,n*m+1);
                else
                {
                    if(vis[xx][yy]==1)continue;
                    uni((i-1)*m+j,(xx-1)*m+yy);
                }
            }
        }
    }
    int x = fi(0);
    int y = fi(n*m+1);
    if(x==y)return 1;
    return 0;
}
void solve()
{
    memset(vis,0,sizeof(vis));
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
        scanf("%s",s[i]+1);
    scanf("%d",&q);
    for(int i=1;i<=q;i++)
    {
        scanf("%d%d",&d[i].first,&d[i].second);
        d[i].first++,d[i].second++;
    }

    if(check(q)){
        printf("-1\n");
        return;
    }

    int l = 0,r = q,ans = 0;
    while(l<=r)
    {
        int mid = (l+r)/2;
        if(!check(mid))r=mid-1,ans=mid;
        else l=mid+1;
    }
    printf("%d\n",ans);
}
int main()
{
    int t;scanf("%d",&t);
    while(t--)solve();
}

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