POJ 2029 Get Many Persimmon Trees DP/二维树状数组/线段树

题意：给定一个由坐标系构成的柿子树林，有的坐标点上存在柿子树，有的不存在。现在准备送给你一块土地，土地的长和宽已经确定，怎么选才能使给你的土地上的柿子树最多。
题解：这道题和 2482 Stars in Your Window 颇为相似。下面给出了三种解法。

方法一：DP

#include <iostream>using namespace std;#define N 103int sum[N][N];bool flag[N][N];int main(){int n, w, h, s, t, x, y, temp, max, i, j;while ( scanf("%d",&n) && n ){memset(flag,0,sizeof(flag));scanf("%d%d",&w,&h);for ( i = 1; i <= n; i++ ){scanf("%d%d",&x,&y);flag[x][y] = true;}scanf("%d%d",&s,&t);sum[0][0] = 0;for ( i = 1; i <= w; i++ ){for ( j = 1; j <= h; j++ )sum[i][j] = sum[i-1][j] + sum[i][j-1] - sum[i-1][j-1] + flag[i][j];}max = 0;for ( i = s; i <= w; i++ ){for ( j = t; j <= h; j++ ){temp = sum[i][j] - sum[i-s][j] - sum[i][j-t] + sum[i-s][j-t];if ( temp > max ) max = temp;}}printf("%d\n", max );}return 0;}

方法二：树状数组

#include<iostream>using namespace std;#define N 105#define max(a,b) ( a > b ? a : b )int c[N][N];int n, w, h, x, y, s, t;int lowbit (int t){return t & ( -t );}int sum ( int x, int y ){int i, j, cnt = 0;for ( i = x; i >= 1; i -= lowbit(i) ){for ( j = y; j >= 1; j -= lowbit(j) )cnt += c[i][j];}return cnt;}void add ( int x, int y ){int i, j;for ( i = x; i <= w; i += lowbit(i) ){for ( j = y; j <= h; j += lowbit(j) )c[i][j]++;}}int get_sum ( int x1, int y1, int x2, int y2 ){    return sum(x2,y2) - sum(x1,y2) - sum(x2,y1) + sum(x1,y1);}int main(){int ans, i, j;while ( scanf("%d",&n) && n ){memset ( c, 0, sizeof(c) );scanf("%d%d",&w,&h);while ( n-- ){scanf("%d%d", &x, &y );add ( x, y );}scanf("%d%d", &s, &t);ans = 0;for( i = s; i <= w; i++ ){for ( j = t; j <= h; j++ )ans = max ( ans, get_sum ( i-s, j-t, i, j ) );}printf("%d\n",ans);}return 0;}

方法三：线段树

#include <algorithm>#include <iostream>using namespace std;#define N 1002#define L(u) ( u * 2 )#define R(u) ( u * 2 + 1 )#define max(a,b) ( a > b ? a : b )int Y[N];struct TreeNode{int l, r, v, add;} node[N*3];struct Point{int x, y, v;} point[N];bool cmp ( Point & a, Point & b ){if ( a.x == b.x )return a.v < b.v; /* 边界注意，当 x 一样时，v 为负值的在前面 ! */return a.x < b.x;}int bfind ( int l, int r, int key ){while ( l <= r ){int mid = ( l + r ) / 2;if ( Y[mid] == key )return mid;if ( key < Y[mid] )r = mid - 1;elsel = mid + 1;}return -1;}void build ( int u, int l, int r ){node[u].l = l;node[u].r = r;node[u].v = node[u].add = 0;if ( l == r ) return;int mid = ( l + r ) / 2;build ( L(u), l, mid );build ( R(u), mid + 1, r );}void update ( int u, int l, int r, int val ){if ( l <= node[u].l && node[u].r <= r ){node[u].add += val;node[u].v += val;return;}if ( node[u].add != 0 ){node[L(u)].add += node[u].add;node[L(u)].v += node[u].add;node[R(u)].add += node[u].add;node[R(u)].v += node[u].add;node[u].add = 0;}int mid = ( node[u].l + node[u].r ) / 2;    if ( r <= mid )     update ( L(u), l, r, val );else if ( l > mid )update ( R(u), l, r, val );else{update ( L(u), l, mid, val );update ( R(u), mid + 1, r, val );}node[u].v = max ( node[L(u)].v, node[R(u)].v );}int main(){int n, w, h, s, t, x, y, i, j;while ( scanf("%d",&n) && n ){scanf("%d%d",&w,&h);for ( i = 1; i <= n; i++ ){scanf("%d%d",&x,&y);point[i].x = x;point[i].y = y;point[i].v = 1;Y[i] = y;}scanf("%d%d",&s,&t);for ( i = 1; i <= n; i++ ){point[i+n].x = point[i].x + s;point[i+n].y = point[i].y;point[i+n].v = -1;Y[i+n] = point[i].y + t;}sort ( point+1, point+1+n*2, cmp );sort ( Y+1, Y+1+n*2 );for ( i = j = 1; i <= n*2; i++ ){if ( Y[j] != Y[i] )Y[++j] = Y[i];}build ( 1, 1, j );   int l, r, ans = 0;  for ( i = 1; i <= n * 2; i++ ){l = bfind ( 1, j, point[i].y );r = bfind ( 1, j, point[i].y + t ) - 1;update ( 1, l, r, point[i].v );if ( ans < node[1].v ) ans = node[1].v;}printf("%d\n", ans );}return 0;}