一个多月前在成电OJ上看到一个并查集的问题，当时刚看完一些基础的数据结构，所以就试着做了。当然没有做出来。这题坑就在于：每组测试数据有10^5个，地图还是1000*1000的。挨着个每次都搜一遍不太可能。容易超时。
解决方法：潮水高度是递增的。所以想象一个退潮的过程，把每次退潮露出的岛屿记录下来，从最后最高的潮水高度算起。把露出的岛屿加入集合，每个测试数据对应的答案可以接着在下一组继续使用。
Islands
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 [1,10^9] separated by single spaces, denoting the heights of the respective fields. Next line contains an integer T (1≤T≤105). The last line contains T integers tj , separated by single spaces, such that 0≤t1≤t2≤⋯≤tT≤10^9
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
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
#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
class Field{
public:
int height, father;
bool under;
};
int Z; int row, col, T;
int res[100000];
vector<int> height_list[100000];
vector<int> t;
int i,j,result,temp;
Field field[1000][1000];
int Father(int a, int b){
if (field[a][b].father!=a*col+b)
field[a][b].father=Father(field[a][b].father/col, field[a][b].father%col);
return field[a][b].father;
}
void Merge(int a, int b, int c, int d){
int f1, f2, x, y;
f1=Father(a, b); f2=Father(c, d);
if(f1==f2) return;
else{
result--;
x=field[a][b].father/col;
y=field[a][b].father%col;
field[x][y].father=f2;
}
}
void Check(int r, int c){
if (r+1<row && !field[r+1][c].under) {Merge(r, c, r+1, c); }
if (r-1>=0 && !field[r-1][c].under) {Merge(r, c, r-1, c); }
if (c+1<col && !field[r][c+1].under) {Merge(r, c, r, c+1); }
if (c-1>=0 && !field[r][c-1].under) {Merge(r, c, r, c-1); }
}
void Solve(){
int x, y, m, n;
result=0;
for(m=t.size()-1; m>=0; m--){
for(n=height_list[m].size()-1; n>=0; n--){
x=height_list[m][n]/col; y=height_list[m][n]%col;
field[x][y].under=false;
result++; Check(x, y);
}
res[m]=result;
}
}
int main(){
scanf("%d", &Z);
while(Z--){
result=0;
t.clear();
for (i=0; i<100000; i++) height_list[i].clear();
scanf("%d%d", &row, &col);
for(i=0; i<row; i++){
for(j=0; j<col; j++){
scanf("%d", &field[i][j].height);
field[i][j].under=true;
field[i][j].father=i*col+j;
}
}
scanf("%d", &T);
for(i=0; i<T; i++){
scanf("%d", &temp);
t.push_back(temp);
}
for(i=0; i<row; i++){
for(j=0; j<col; j++){
temp=lower_bound(t.begin(), t.end(), field[i][j].height)-t.begin();
if(temp>0) height_list[temp-1].push_back(i*col+j);
}
}
Solve();
for(i=0; i<(int)t.size(); i++) printf("%d ", res[i]);
printf("\n");
}
return 0;
}