POJ 1195 2维线段树(树套树实现) 树状数组

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <iostream>
using namespace std;

#define LL(a)   a<<1
#define RR(a)   a<<1|1
const int MaxL = 1005;
struct sub_Seg_tree_node
{
    int subleft,subright; // ydown = subleft, yup = subright
//    int maxval;  // the maval of the rect x [left,right], y [subleft, subright]
    int sum;     // the sum of the num of point in rect x [left, right], y [subleft, subright]
};

struct Seg_tree_node
{
    int left,right;       // left = xleft, right = xright
    sub_Seg_tree_node T[MaxL<<2];
} TT[MaxL<<2];

void sub_build(int subl, int subr, int subidx, int idx)
{
    TT[idx].T[subidx].subleft=subl;
    TT[idx].T[subidx].subright=subr;
//    TT[idx].T[subidx].maxval=-1;
    TT[idx].T[subidx].sum=0;
    if(subl==subr)  return;
    int mid=(subl+subr)>>1;
    sub_build(subl, mid, LL(subidx), idx);
    sub_build(mid+1, subr, RR(subidx), idx);
}

void build(int l, int r, int subl, int subr, int idx)
{
    TT[idx].left = l, TT[idx].right = r;
    sub_build(subl, subr, 1, idx);
    if(l==r) return;
    int mid=(l+r)>>1;
    build(l,mid,subl,subr, LL(idx));
    build(mid+1,r,subl,subr, RR(idx));
}

void sub_update(int y, int val, int subidx, int idx)
{
    TT[idx].T[subidx].sum += val;
//    TT[idx].T[subidx].maxval=max(TT[idx].T[subidx].maxval, val);
    if(TT[idx].T[subidx].subleft == TT[idx].T[subidx].subright) return;
    int mid=(TT[idx].T[subidx].subleft+TT[idx].T[subidx].subright)>>1;
    if(y <= mid)
        sub_update(y, val, LL(subidx), idx);
    else
        sub_update(y, val, RR(subidx), idx);
}

void update(int x, int y, int val, int idx)
{
    sub_update(y, val, 1, idx);
    if(TT[idx].left == TT[idx].right) return;
    int mid=(TT[idx].left+TT[idx].right)>>1;
    if( x<=mid)
        update(x,y,val, LL(idx) );
    else
        update(x,y, val,RR(idx));
}

//int sub_query(int subl, int subr, int subidx, int idx)
//{
//    if(TT[idx].T[subidx].subleft == subl && TT[idx].T[subidx].subright == subr)
//        return TT[idx].T[subidx].maxval;
//    int mid = (TT[idx].T[subidx].subleft + TT[idx].T[subidx].subright)>>1;
//    if(subr <= mid)
//        return sub_query(subl, subr, LL(subidx), idx);
//    else if(subl > mid)
//        return sub_query(subl, subr, RR(subidx), idx);
//    else
//        return max(sub_query(subl, mid, LL(subidx), idx),sub_query(mid+1, subr, RR(subidx), idx));
//}
//
//int query(int l, int r, int subl, int subr, int idx)
//{
//    if(TT[idx].left == l && TT[idx].right == r)
//        return sub_query(subl, subr, 1, idx);
//    int mid = (TT[idx].left + TT[idx].right)>>1;
//    if(r <= mid)
//        return query(l,r, subl,subr, LL(idx));
//    else if(l > mid)
//        return query(l, r, subl, subr, RR(idx));
//    else
//        return max(query(l, mid, subl, subr, LL(idx)),query(mid+1, r, subl,subr, RR(idx)));
//}

int sub_query(int subl, int subr, int subidx, int idx)
{
    if(TT[idx].T[subidx].subleft == subl && TT[idx].T[subidx].subright == subr)
        return TT[idx].T[subidx].sum;
    int mid = (TT[idx].T[subidx].subleft + TT[idx].T[subidx].subright)>>1;
    if(subr <= mid)
        return sub_query(subl, subr, LL(subidx), idx);
    else if(subl > mid)
        return sub_query(subl, subr, RR(subidx), idx);
    else
        return sub_query(subl, mid, LL(subidx), idx) + sub_query(mid+1, subr, RR(subidx), idx);
}

int query(int l, int r, int subl, int subr, int idx)
{
    if(TT[idx].left == l && TT[idx].right == r)
        return sub_query(subl, subr, 1, idx);
    int mid = (TT[idx].left + TT[idx].right)>>1;
    if(r <= mid)
        return query(l,r, subl,subr, LL(idx));
    else if(l > mid)
        return query(l, r, subl, subr, RR(idx));
    else
        return query(l, mid, subl, subr, LL(idx))+ query(mid+1, r, subl,subr, RR(idx));
}

int x1[10005];
int y1[10005];
int x2[10005];
int y2[10005];
int px[10005];
int py[10005];
int pval[10005];
//
//int main()
//{
//    freopen("1.txt","r",stdin);
//    int NN,MM;
//    cin>>NN>>MM;
//
//    for(int i=0; i<NN; i++)
//        scanf("%d%d%d", &px[i],&py[i], &pval[i]);
//
//    for(int i=0; i<MM; i++)
//    {
//        scanf("%d%d%d%d", &x1[i], &y1[i], &x2[i], &y2[i]);
//        if(x1[i] > x2[i]) swap(x1[i], x2[i]);
//        if(y1[i] > y2[i]) swap(y1[i], y2[i]);
//    }
//
//    build(0, 1060, 0, 1060, 1);
//    for(int i=0; i<NN; i++)
//    {
//        cout<<"update point "<<px[i]<<" "<<py[i]<<endl;
//        update(px[i],py[i],pval[i], 1);
//    }
//
//    for(int i=0; i<MM; i++)
//    {
//        cout<<"query ret "<<x1[i]<<" "<<x2[i]<<" "<<y1[i]<<" "<<y2[i]<<endl;
//        cout<<"Ret "<<query(x1[i],x2[i],y1[i],y2[i],1)<<endl;
//    }
//
//    return 0;
//}
int s;
int main()
{
//    freopen("1.txt","r",stdin);
    int i;
    int sb;
    int sb1,sb2,sb3,sb4;
    scanf("%d",&sb);
    while(sb!=3)
    {
        switch(sb)
        {
        case 0:
            scanf("%d",&s);
            build(0,s,0,s,1);
            break;
        case 1:
            scanf("%d%d%d",&sb1,&sb2,&sb3);
            update(sb1,sb2,sb3,1);
            break;
        case 2:
            scanf("%d%d%d%d",&sb1,&sb2,&sb3,&sb4);
            printf("%d
",query(sb1,sb3,sb2,sb4,1));
            break;
        default:
            goto ed;
        }
        scanf("%d",&sb);
    }
ed:
    ;
    return 0;
}

树状数组来做这个题目相对来说简单得多!  树状数组则清楚得多.相对来说二维线段树用得比较少,而且功能都比较简单,查询平面矩形包含几个点,查询矩形最大值之类的.

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;

#define MAX 1026
int s[MAX][MAX];
int lowbit(int x)
{
    return x&(x^(x-1));
}

int N;

void change(int a,int b,int delta)
{
    for(int x=a; x<=N; x+=lowbit(x))
        for(int y=b; y<=N; y+=lowbit(y))
            s[x][y]+=delta;
}
int sum(int a,int b)
{
    int res=0;
    for(int x=a; x>0; x-=lowbit(x))
        for(int y=b; y>0; y-=lowbit(y))
            res+=s[x][y];
    return res;
}
int main()
{
    int n;
    while(cin>>n)
    {
        if(n==0)
        {
            cin>>N; N++;
            memset(s,0,sizeof(s));
        }
        else if(n==1)
        {
            int a,b,c;
            cin>>a>>b>>c; a++; b++;
            change(a,b,c);
        }
        else if(n==2)
        {
            int a,b,c,d;
            cin>>a>>b>>c>>d;
            a++;b++;c++;d++;
            cout<<sum(c,d) + sum(a-1,b-1) - sum(a-1,d) - sum(c,b-1)<<endl;
        }
        else break;;
    }
    return 0;
}