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Problem H

Time Limit : 5000/3000ms (Java/Other)   Memory Limit : 65535/32768K (Java/Other)

Total Submission(s) : 8   Accepted Submission(s) : 3

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Problem Description

Input

Output

Sample Input

2
3
S 1 1 1 1
A 1 1 2
S 1 1 1 1
3
S 1 1 1 1
A 1 1 2
S 1 1 1 2

Sample Output

Case 1:
1
3
Case 2:
1
4

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int n=1010;
int c[n+1][n+1];
int lowbit(int x)
{
    return x&(-x);
}
void update(int x,int y,int val)
{
    for(int i=x;i<=n;i+=lowbit(i))
    {
        for(int j=y;j<=n;j+=lowbit(j))
        {
            c[i][j]+=val;
        }
    }
}
int getsum(int x,int y)
{
    int cnt=0;
    for(int i=x;i>=1;i-=lowbit(i))
    {
        for(int j=y;j>=1;j-=lowbit(j))
        {
            cnt+=c[i][j];
        }
    }
    return cnt;
}
int val[n+1][n+1];
int main()
{
    int ci;scanf("%d",&ci);
    int pl=1;
    while(ci--)
    {
        memset(c,0,sizeof(c));
        printf("Case %d:\n",pl++);
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                update(i,j,1);
                val[i][j]=1;
            }
        }
        int sa;scanf("%d",&sa);
        while(sa--)
        {
            char ch;cin>>ch;
            if(ch==‘S‘)
            {
                int xx1,yy1,xx2,yy2;
                scanf("%d%d%d%d",&xx1,&yy1,&xx2,&yy2);
                xx1++,xx2++,yy1++,yy2++;//从1开始
                int x1,x2,y1,y2;
                x1=min(xx1,xx2);x2=max(xx1,xx2);
                y1=min(yy1,yy2);y2=max(yy1,yy2);
                int cnt=getsum(x2,y2)-getsum(x1-1,y2)-getsum(x2,y1-1)+getsum(x1-1,y1-1);
                printf("%d\n",cnt);
            }
            else if(ch==‘A‘)
            {
                int x,y,l;
                scanf("%d%d%d",&x,&y,&l);
                x++,y++;
                update(x,y,l);
                val[x][y]+=l;
            }
            else if(ch==‘D‘)
            {
                int x,y,l;
                scanf("%d%d%d",&x,&y,&l);
                x++,y++;
                if(l>val[x][y]) l=val[x][y];//important
                update(x,y,-l);
                val[x][y]+=-l;
            }
            else
            {
                int x1,x2,y1,y2,l;
                scanf("%d%d%d%d%d",&x1,&y1,&x2,&y2,&l);
                x1++,x2++,y1++,y2++;
                if(l>val[x1][y1]) l=val[x1][y1];//important
                update(x1,y1,-l);
                val[x1][y1]+=-l;
                update(x2,y2,l);
                val[x2][y2]+=l;
            }
        }
    }
    return 0;
}

hdoj 1892(二维树状数组),布布扣,bubuko.com

hdoj 1892(二维树状数组)

标签：des   java   os   io   for   ar   div   代码   

原文地址：http://www.cnblogs.com/weiyikang/p/3900651.html