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 n×m 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 4×5 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.

Input

Multiple Test Cases

The input contains several test cases. The first line of the input contains a positive integer Z≤20,denoting the number of test cases. Then Z test cases follow, each conforming to the format described in section Single Instance Input. For each test case, your program has to write an output conforming to the format described in section Single Instance Output.

Single Instance Input

The first line contains two numbers n and m separated by a single space, the dimensions of the island, where 1≤n,m≤1000. Next n lines 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≤T≤105). The last line contains Tintegers tj , separated by single spaces, such that 0≤t1≤t2≤?≤tT≤109

Output

Single Instance Output

Your program should output a single line consisting of T numbers rj , where rj is the number of unflooded areas in year tj . After every number ,you must output a single space!

Sample input and output

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

2 3 1 0 0

题意就是有一个矩阵，求第i天矩阵中大于i的部分连通块的个数。

题目会给出i，i是递增的，所以可以从最后一天做起，每天被昨天淹没且不被今天淹没的大陆浮起来，并将其与已经浮起来的大陆连接起来，再求连通块的个数。连接这里需要用到并查集，而由于i最大是10^9，但i的询问个数不超过10000，所以要离散化。

这道题还有个坑就是时间限制居然是10s，一开始怎么想都是超时=。=

#include <iostream>
#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<vector>
using namespace std;
int q[1005][1005];
vector<int> w;
vector<int> x[100005];
int found[1005][1005];
int father[1005][1005];
int hang,lie,ans;
int Rank[1100005]={};
int getfather(int j,int k)
{
    if (father[j][k]!=j*1000+k-1)
        father[j][k]=getfather(father[j][k]/1000,father[j][k]%1000+1);
    return father[j][k];
}
void judge(int g,int h,int j,int k)
{
    int fx,fy;
    if ((fx=getfather(g,h))==(fy=getfather(j,k))) return ;
    ans--;
    if (Rank[fx]>Rank[fy]) father[fy/1000][fy%1000+1]=fx;
    else{
            father[fx/1000][fx%1000+1]=fy;
            if (Rank[fx]==Rank[fy])
                Rank[fx]++;
        }
    return;
}
void dfs(int g,int h)
{
    if (found[g][h]) return;
    found[g][h]++;
    ans++;
    if (found[g+1][h] && g<hang) judge(g,h,g+1,h);
    if (found[g-1][h] && g>1) judge(g,h,g-1,h);
    if (found[g][h+1] && h<lie) judge(g,h,g,h+1);
    if (found[g][h-1] && h>1) judge(g,h,g,h-1);
    return ;
}
int main()
{
    int g,h,j,k,l,n,T;
    cin>>T;
    while (T--){
        memset(found,0,sizeof(found));
        cin>>g>>h;
        memset(q,0,sizeof(q));
        for (j=1;j<=g;j++)
            for(k=1;k<=h;k++)
                scanf("%d",&q[j][k]);
        cin>>l;
        hang=g;
        lie=h;
        for (j=1;j<=g;j++) for (k=1;k<=h;k++) father[j][k]=j*1000+k-1;
        ans=0;
        memset(Rank,0,sizeof(Rank));
        vector <int>().swap(w);
        for (j=0;j<100005;j++) vector <int>().swap(x[j]);
        w.push_back(0);
        for (j=1;j<=l;j++){
            scanf("%d",&n);
            w.push_back(n);
        }
        w.push_back(1000000007);
        for (j=1;j<=g;j++)
            for(k=1;k<=h;k++)
                q[j][k]=lower_bound(w.begin(),w.end(),q[j][k])-w.begin();
        for (j=1;j<=g;j++)
            for (k=1;k<=h;k++)
                if (q[j][k]>=0) x[q[j][k]].push_back(j*1000+k-1);
        for (j=w.size()-2;j>0;j--){
            for (int k=0;k<x[j+1].size();k++)
                dfs((int)x[j+1][k]/1000,(int)x[j+1][k]%1000+1);
            w[j]=ans;
        }
        for (j=1;j<w.size()-1;j++) printf("%d ",w[j]);
        cout<<endl;
    }
    return 0;
}

这道题代码不是很长，我居然前后用了6个小时才AC，看来是我太弱了。。。。。