用水填坑 （牛客）

https://ac.nowcoder.com/acm/contest/403/A

解析：該題很明顯是一道搜索題。首先我們必須明確在什麼情況下會形成水窪：即當某一點的高度小於等於( <= )上下左右四個方向時，會形成水窪。

           該題是要計算形成水窪的面積，那麼我們的某一塊最大存水量一定是四個方向上最小的高度。

 

1. 第一次的寫法：只是使用深搜去枚舉每一個點，但寫完發現這樣寫是有問題，對每一點枚舉時，只是一直在對周圍比較，卻沒有考慮離該點更遠的點，這樣顯然時不對的

wa的代碼：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<string>

using namespace std;

const int maxn = 1e3+5;
const int INF = 1e9+7;

int a[maxn][maxn];
int h[maxn][maxn];
int vis[maxn][maxn];
int n, m;

int xx[] = {0, -1, 0, 1};
int yy[] = {-1, 0, 1, 0};

void dfs(int x, int y){
	//printf("%d  %d\n", x, y);
	if(x == n && y == m){
	//	printf("%d %d\n", x, y);
		return ;
	}
	if(y <= m) dfs(x, y+1);
	int num = 0, mi = INF;
	if(!vis[x][y]){
		for(int i = 0; i < 4; i++){
			int dx, dy; 
			dx = x + xx[i];
			dy = y + yy[i];
			if(dx >= 1 && dx <= n && dy >= 1 && dy <= m){
				if(a[x][y] <= h[dx][dy]){
					num++;
					mi = min(mi, h[dx][dy]);
				}
			}
		}
		//printf("!!!!! %d %d\n", mi, num);
	}
	if(num == 4){
		h[x][y] = mi;
	}
	vis[x][y] = 1;
	if(x <= n) dfs(x+1, y);
}

int main()
{
	scanf("%d%d", &n, &m);
	for(int i = 1; i <= n; i++)
		for(int j = 1; j <= m; j++){
			scanf("%d", &a[i][j]);
			h[i][j] = a[i][j];
			if(i == 1 || i == n || j == 1 || j == m)
				vis[i][j] = 1;
		} 
	
	int ans = 0;
	dfs(1, 1);
	for(int i = 1; i <= n; i++){
		for(int j = 1; j <= m; j++){
			printf("%d ", h[i][j]);
		}
		printf("\n");
	}
	printf("%d\n", ans);
	return 0;
}

/*
0 0 0 0 0 0 0 0 0 0
0 5 2 6 4 3 1 7 8 0
0 6 4 2 3 5 1 4 6 0
0 3 6 4 1 2 4 7 8 0
0 2 5 5 1 2 3 4 4 0
0 2 3 1 5 4 4 1 4 0 
0 4 1 2 3 4 5 2 1 0
0 7 5 5 1 5 4 5 7 0
0 1 3 5 5 4 6 8 7 0
0 0 0 0 0 0 0 0 0 0


0 0 0 0 0 0 0 0 0 0
0 5 2 6 4 3 1 7 8 0
0 6 4 3 3 5 1 4 6 0
0 3 6 4 1 2 4 7 8 0
0 2 5 5 1 2 3 4 4 0
0 2 3 2 5 4 4 2 4 0
0 4 2 2 3 4 5 2 1 0
0 7 5 5 3 5 5 5 7 0
0 1 3 5 5 4 6 8 7 0
0 0 0 0 0 0 0 0 0 0
*/

1. 使用已經確定的點來不斷更新未確定的點，知道遍歷所有的點。在這裏使用堆來進行維護，可以更加方便。ac

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<string>
#include<queue>

using namespace std;

const int maxn = 1e3+5;

struct Node{
	int x, y, h;
	bool operator < (const Node &a) const 
	{
		return h > a.h;
	}
};

priority_queue<Node>pq;
int a[maxn][maxn];
int h[maxn][maxn];
int vis[maxn][maxn];

int n, m;

int xx[] = {0, -1, 0, 1};
int yy[] = {-1, 0, 1, 0};

int main()
{
	scanf("%d%d", &n, &m);
	for(int i = 1; i <= n; i++){
		for(int j = 1; j <= m; j++){
			scanf("%d", &a[i][j]);
			h[i][j] = a[i][j];
			if(i == 1 || i == n || j ==1 || j == m){
				vis[i][j] = 1;
				pq.push(Node{i, j, a[i][j]});
			}
		}
	}
	
	while(!pq.empty()){
		Node t = pq.top(); pq.pop();
		int tmph = t.h;
		for(int i = 0; i < 4; i++){
			int dx = t.x + xx[i];
			int dy = t.y + yy[i];
			if(dx < 1 || dx > n || dy < 1 || dy > m || vis[dx][dy]) continue;
			h[dx][dy] = max(h[dx][dy], tmph);
			vis[dx][dy] = 1;
			pq.push(Node{dx, dy, h[dx][dy]});
		}
	}
	
	long long ans = 0;
	for(int i = 1; i <= n; i++){
		for(int j = 1; j <= m; j++){
			//ans += (h[i][j] - a[i][j]) * 1LL;
			printf("%d ", h[i][j]);
		}
		printf("\n");	
	}
	printf("%lld\n", ans);
	return 0;
}
/*
0 0 0 0 0 0 0 0 0 0
0 5 2 6 4 3 1 7 8 0
0 6 4 2 3 5 1 4 6 0
0 3 6 4 1 2 4 7 8 0
0 2 5 5 1 2 3 4 4 0
0 2 3 1 5 4 4 1 4 0 
0 4 1 2 3 4 5 2 1 0
0 7 5 5 1 5 4 5 7 0
0 1 3 5 5 4 6 8 7 0
0 0 0 0 0 0 0 0 0 0


0 0 0 0 0 0 0 0 0 0
0 5 2 6 4 3 1 7 8 0
0 6 4 4 4 5 1 4 6 0
0 3 6 4 4 4 4 7 8 0
0 2 5 5 4 4 4 4 4 0
0 2 3 3 5 4 4 2 4 0
0 4 3 3 3 4 5 2 1 0
0 7 5 5 3 5 5 5 7 0
0 1 3 5 5 4 6 8 7 0
0 0 0 0 0 0 0 0 0 0
*/