AtCoder Grand Contest 015 C - Nuske vs Phantom Thnook

题目传送门：https://agc015.contest.atcoder.jp/tasks/agc015_c

题目大意：

现有一个\(N×M\)的矩阵\(S\)，若\(S_{i,j}=1\)，则该处为蓝色，否则为白色。保证蓝色格子构成的联通块为树，即联通块内蓝格子相邻的边为\(cnt-1\)，多次询问子矩阵内蓝色联通块个数

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题目都保证是棵树了就是个SB题，二维前缀和，记录蓝块数目，相邻蓝块个数（行列分开记好写些），子矩阵内用蓝块个数减去相邻蓝块个数即为联通块个数

/*program from Wolfycz*/

#include<cmath>

#include<cstdio>

#include<cstring>

#include<iostream>

#include<algorithm>

#define inf 0x7f7f7f7f

using namespace std;

typedef long long ll;

typedef unsigned int ui;

typedef unsigned long long ull;

inline char gc(){

	static char buf[1000000],*p1=buf,*p2=buf;

	return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++;

}

inline int frd(){

	int x=0,f=1; char ch=gc();

	for (;ch<'0'||ch>'9';ch=gc())	if (ch=='-')	f=-1;

	for (;ch>='0'&&ch<='9';ch=gc())	x=(x<<3)+(x<<1)+ch-'0';

	return x*f;

}

inline int read(){

	int x=0,f=1; char ch=getchar();

	for (;ch<'0'||ch>'9';ch=getchar())	if (ch=='-')	f=-1;

	for (;ch>='0'&&ch<='9';ch=getchar())	x=(x<<3)+(x<<1)+ch-'0';

	return x*f;

}

inline void print(int x){

	if (x<0)	putchar('-'),x=-x;

	if (x>9)	print(x/10);

	putchar(x%10+'0');

}

const int N=2e3;

char map[N+10][N+10];

int sum[N+10][N+10],lne[N+10][N+10],row[N+10][N+10];

int main(){

	int n=read(),m=read(),q=read();

	for (int i=1;i<=n;i++){

		scanf("%s",map[i]+1);

		for (int j=1;j<=m;j++){

			if (map[i][j]=='1'){

				sum[i][j]=1;

				if (map[i][j-1]=='1')	lne[i][j]=1;

				if (map[i-1][j]=='1')	row[i][j]=1;

			}

		}

	}

	for (int i=1;i<=n;i++){

		for (int j=1;j<=m;j++){

			sum[i][j]+=sum[i-1][j]+sum[i][j-1]-sum[i-1][j-1];

			lne[i][j]+=lne[i-1][j]+lne[i][j-1]-lne[i-1][j-1];

			row[i][j]+=row[i-1][j]+row[i][j-1]-row[i-1][j-1];

		}

	}

	for (int i=1;i<=q;i++){

		int x1=read(),y1=read(),x2=read(),y2=read(),Ans=0;

		Ans+=sum[x2][y2]-sum[x1-1][y2]-sum[x2][y1-1]+sum[x1-1][y1-1];

		Ans-=lne[x2][y2]-lne[x1-1][y2]-lne[x2][y1]+lne[x1-1][y1];

		Ans-=row[x2][y2]-row[x1][y2]-row[x2][y1-1]+row[x1][y1-1];

		printf("%d\n",Ans);

	}

	return 0;

}