hdu_2642 Stars 线段树+点修改

Stars

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/65536 K (Java/Others)
Total Submission(s): 1196    Accepted Submission(s): 485

Problem Description

Yifenfei is a romantic guy and he likes to count the stars in the sky.
To make the problem easier,we considerate the sky is a two-dimension plane.Sometimes the star will be bright and sometimes the star will be dim.At first,there is no bright star in the sky,then some information will be given as "B x y" where 'B' represent bright and x represent the X coordinate and y represent the Y coordinate means the star at (x,y) is bright,And the 'D' in "D x y" mean the star at(x,y) is dim.When get a query as "Q X1 X2 Y1 Y2",you should tell Yifenfei how many bright stars there are in the region correspond X1,X2,Y1,Y2.

There is only one case.

 

 

Input

The first line contain a M(M <= 100000), then M line followed.
each line start with a operational character.
if the character is B or D,then two integer X,Y (0 <=X,Y<= 1000)followed.
if the character is Q then four integer X1,X2,Y1,Y2(0 <=X1,X2,Y1,Y2<= 1000) followed.

 

 

Output

For each query,output the number of bright stars in one line.

 

 

Sample Input

5B 581 145B 581 145Q 0 600 0 200D 581 145Q 0 600 0 200

 

 

Sample Output

10

 

#include <map>#include <set>#include <stack>#include <queue>#include <cmath>#include <vector>#include <cstdio>#include <cstring>#include <cstdlib>#include <iostream>#include <algorithm>using namespace std;#define INF 0x3f3f3f3f#define inf -0x3f3f3f3f#define FOR(a,b) for(int i =  a ; i < b ; i++)#define mem0(a) memset(a,0,sizeof(a))#define mem1(a) memset(a,-1,sizeof(a))#define FOPENIN(IN) freopen(IN, "r", stdin)#define FOPENOUT(OUT) freopen(OUT, "w", stdout)const int num = 1000+10;int M;int c[num][num],a[num][num];int lowbit(int x){    return x&(-x);}void add(int x,int y,int k){    for(int i = x ; i > 0 ; i-=lowbit(i))        for(int j = y ; j > 0 ; j-=lowbit(j))           c[i][j]+=k;}int Getsum(int x,int y){    int ans = 0 ;    for(int i = x ; i < num ; i+= lowbit(i))        for(int j = y ; j < num ; j+= lowbit(j)){            ans += c[i][j];    }    return ans;}int main(){    while(scanf("%d",&M)!=EOF){        mem0(a);mem0(c);        while(M--){            char cc;            int x1,y1,x2,y2;            cin>>cc;            if(cc=='B'){                scanf("%d%d",&x1,&y1);                x1++;y1++;                if(a[x1][y1])continue;                a[x1][y1]=1;                add(x1,y1,1);//                printf("\n******\n");            }            else if(cc=='D'){                scanf("%d%d",&x1,&y1);                x1++;y1++;                if(a[x1][y1]==0)continue;                a[x1][y1]=0;                add(x1,y1,-1);            }            else if(cc=='Q'){                scanf("%d%d%d%d",&x1,&x2,&y1,&y2);                x1++;y1++;x2++;y2++;                if(x1>x2)swap(x1,x2);                if(y1>y2)swap(y1,y2);                int Sum=Getsum(x1,y1)-Getsum(x2+1,y1)-Getsum(x1,y2+1)+Getsum(x2+1,y2+1);                printf("%d\n",Sum);            }        }    }    return 0;}