題目連結

時間限制：C/C++ 4秒，其他語言8秒
空間限制：C/C++ 262144K，其他語言524288K
64bit IO Format: %lld

題目描述

White Rabbit has a rectangular farmland of n*m. In each of the grid there is a kind of plant. The plant in the j-th column of the i-th row belongs the a[i][j]-th type.
White Cloud wants to help White Rabbit fertilize plants, but the i-th plant can only adapt to the i-th fertilizer. If the j-th fertilizer is applied to the i-th plant (i!=j), the plant will immediately die.
Now White Cloud plans to apply fertilizers T times. In the i-th plan, White Cloud will use k[i]-th fertilizer to fertilize all the plants in a rectangle [x1[i]...x2[i]][y1[i]...y2[i]].
White rabbits wants to know how many plants would eventually die if they were to be fertilized according to the expected schedule of White Cloud.

輸入描述:

The first line of input contains 3 integers n,m,T(n*m<=1000000,T<=1000000)
For the next n lines, each line contains m integers in range[1,n*m] denoting the type of plant in each grid.
For the next T lines, the i-th line contains 5 integers x1,y1,x2,y2,k(1<=x1<=x2<=n,1<=y1<=y2<=m,1<=k<=n*m)
 

輸出描述:

Print an integer, denoting the number of plants which would die.

示例1

輸入

2 2 2

1 2

2 3

1 1 2 2 2

2 1 2 1 1

輸出

3

題意:給出一個N*M的矩陣藥田,藥田中每個格子都種有一種編號的為1~N*M的藥,隨後一個T表示有T次澆水操作,之後先是給出N行M列的矩陣(藥田),再給出T行,每行有5個數,X1,Y1,X2,Y2和W,表示這次操作會在左上角為(X1,Y1),右下角為(X2,Y2)的矩陣中澆W這種藥水,若是該某個藥格中的藥編號與被澆到的藥水不同,則死亡,問你T次澆水操作後整個藥田死了多少顆藥?

題解:這裡對每個藥的編號進行隨機hash得到一個大於最大編號1e6的數值,之後每次澆水操作,我們通過使用二維樹狀陣列區間更新的巧妙辦法,在已知更新的矩陣左上角和右下角情況下update四個點的值即可在O(logn*logm)時間內實現樹狀陣列的區間更新(需要注意的是這裡更新的值也是藥對應藥水的hash值才行).那麼最終想要知道某個格子的藥是否存活只需要判斷他的樹狀陣列的值%本身的hash值是否為0即可,在隨機數足夠優秀的條件下,能整除的情況說明該藥格每次澆到的藥水的是合適的(或者沒有澆到任何藥水),那麼它是存活的,反之死亡.

程式碼如下:

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<ctime>
#include<string>
#include<cstring>
#include<vector>
using namespace std;
typedef long long ll;
const int maxn = 1e6+10;
vector<ll>mp[maxn];     //存二維矩陣
vector<ll>tre[maxn];    //存二維樹狀陣列結果
ll has[maxn];
int n, m;
void fun() {           //預處理出1e6以內的所有數的has值(這裡指隨機生成的大於1e6的數)
	srand(time(NULL));
	for (int i = 1; i < maxn - 5; i++)
		has[i] = (ll)rand()*1e6 + rand()*rand();
}
//一下三個函式使用的二維樹狀陣列
ll lowbit(ll x) {     
	return x&(-x);
}
void update(int x, int y, ll z)
{
	for (int i = x; i <= n; i += lowbit(i))
		for (int j = y; j <= m; j += lowbit(j))
			tre[i][j] += z;
}
ll sum(int x, int y)
{
	ll ret = 0;
	for (int i = x; i >= 1; i -= lowbit(i))
		for (int j = y; j >= 1; j -= lowbit(j))
			ret += tre[i][j];
	return ret;
}
int main() {
	int t, x;
	scanf("%d%d%d", &n, &m, &t);
	fun();
	for (int i = 1; i <= n; i++) {
		mp[i].push_back(0);         //使mp的列從1開始(方便後面查詢)
		tre[i].push_back(0);
		for (int j = 1; j <= m; j++) {
			scanf("%d", &x);
			mp[i].push_back(has[x]);//將給出的值進行隨機化
			tre[i].push_back(0);   //初始化樹狀陣列
		}
	}
	int x1, y1, x2, y2;
	ll w;
	while (t--) {                //給的是左上角和右下角的座標
		scanf("%d%d%d%d%lld", &x1, &y1, &x2, &y2, &w);
		update(x1, y1, has[w]);  //通過更新這4個幾個實現區間更新
		update(x2 + 1, y2 + 1, has[w]);
		update(x2 + 1, y1, -has[w]);
		update(x1, y2 + 1, -has[w]);
	}
	int ans = 0;
	for (int i = 1; i <= n; i++) //若是最終在該點撒過得農藥總和值不是原值得倍數,則表示死亡
		for (int j = 1; j <= m; j++)
			if (sum(i, j) % mp[i][j] != 0) ans++;
	printf("%d\n", ans);
	return 0;
}