Round A APAC Test 2017 Problem B.Rain

Problem B. Rain

This contest is open for practice. You can try every problem as many times as you like, though we won't keep track of which problems you solve. Read the Quick-Start
Guide to get started.

Small input 9 points	Solve B-small Judge's response for last submission: Correct.
Large input 15 points	Solve B-large Judge's response for last submission: Correct.

Problem

There's an island in the sea. The island can be described as a matrix with R rows and C columns, with H[i][j] indicating the height of each unit cell. Following is an example of a 3*3 island:

3 5 5
5 4 5
5 5 5

Sometimes, a heavy rain falls evenly on every cell of this island. You can assume that an arbitrarily large amount of water falls. After such a heavy rain, some areas of the island (formed of one or more unit cells joined along edges) might collect water. This
can only happen if, wherever a cell in that area shares an edge (not just a corner) with a cell outside of that area, the cell outside of that area has a larger height. (The surrounding sea counts as an infinite grid of cells with height 0.) Otherwise, water
will always flow away into one or more of the neighboring areas (for our purposes, it doesn't matter which) and eventually out to sea. You may assume that the height of the sea never changes. We will use W[i][j] to denote the heights of the island's cells
after a heavy rain. Here are the heights of the example island after a heavy rain. The cell with initial height 4 only borders cells with higher initial heights, so water will collect in it, raising its height to 5. After that, there are no more areas surrounded
by higher cells, so no more water will collect. Again, note that water cannot flow directly between cells that intersect only at their corners; water must flow along shared edges.

Following is the height of the example island after rain:

3 5 5
5 5 5
5 5 5

Given the matrix of the island, can you calculate the total increased height sum(W[i][j]-H[i][j]) after a heavy rain?

Input

The first line of the input gives the number of test cases, T. T test cases follow.

The first line of each test case contains two numbers R and C indicating the number of rows and columns of cells on the island. Then, there are R lines of C positive integers each. The j-th
value on the i-th of these lines gives H[i][j]: the height of the cell in the i-th row and the j-th column.

Output

For each test case, output one line containing

Case #x: y

, where

x

is the test case number (starting
from 1) and

y

is the total increased height.

Limits

1 ≤ T ≤ 100.

1 ≤ H[i][j] ≤ 1000.

Small dataset

1 ≤ R ≤ 10.

1 ≤ C ≤ 10.

Large dataset

1 ≤ R ≤ 50.

1 ≤ C ≤ 50.

Sample

Input	Output
3 3 3 3 5 5 5 4 5 5 5 54 4 5 5 5 1 5 1 1 5 5 1 5 5 5 2 5 8 4 3 2 2 2 2 1 2 2 1 2 2 1 2	Case #1: 1 Case #2: 3 Case #3: 0

Case 1 is explained in the statement.

In case 2, the island looks like this after the rain:

5 5 5 1
5 2 2 5
5 2 5 5
5 2 5 8

Case 3 remains unchanged after the rain.

虽然是乱搞题。。。但是还有很多细节。。。自己没有想清楚就开写导致浪费了很多时间。。哎。。。

希望明天面试能够脑子清醒，好好表达自己。马友友！！娘娘！！！！give me power！！！

下面代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
using namespace std;
const int maxn = 100;
const int inf = 0x3f3f3f3f;
int save[maxn][maxn];
bool vis[maxn][maxn];
int dx[4] = {0,0,1,-1};
int dy[4] = {1,-1,0,0};
int n,m,sum;
int mx,my;
bool over(){
int i,j;
for(i=1;i<=n;i++){
for(j=1;j<=m;j++){
if(!vis[i][j]){
return true;
}
}
}
return false;
}
bool check(int x,int y){
int i;
for(i=0;i<4;i++){
if(!vis[x+dx[i]][y+dy[i]]){
return true;
}
}
return false;
}
void select(){
int i,j;
for(i=1;i<=n;i++){
for(j=1;j<=m;j++){
if(vis[i][j]&&check(i,j)&&save[mx][my]>save[i][j]){
mx = i;
my = j;
}
}
}
}

void dfs(int x,int y){
int i;
for(i=0;i<4;i++){
int nx = x+dx[i], ny = y + dy[i];
if(nx>=1&&nx<=n&&ny>=1&&ny<=m&&!vis[nx][ny]){
if(save[nx][ny]<=save[mx][my]){
sum += save[mx][my] - save[nx][ny];
save[nx][ny] = save[mx][my];
vis[nx][ny] = true;
dfs(nx,ny);
}else{
vis[nx][ny] = true;
}
}
}
}
int main(){
freopen("in.txt", "r", stdin);
freopen("out.txt","w",stdout);
int t,i,j,rnd = 1;
scanf("%d",&t);
while(t--){
sum = 0;
scanf("%d%d",&n,&m);
memset(vis,true,sizeof(vis));
for(i=1;i<=n;i++){
for(j=1;j<=m;j++){
scanf("%d",&save[i][j]);
vis[i][j] = false;
}
}
for(i=1;i<=n;i++){
vis[i][1] = true;
vis[i][m] = true;
}
for(i=1;i<m;i++){
vis[1][i] = true;
vis
[i] = true;
}
save[0][0] = inf;
while(over()){
mx = 0, my = 0;
select();
dfs(mx,my);
}
printf("Case #%d: %d\n",rnd++,sum);
}
return 0;
}