HDU6183(线段树)

Color it

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 132768/132768 K (Java/Others)
Total Submission(s): 302    Accepted Submission(s): 62

Problem Description

Do you like painting? Little D doesn't like painting, especially messy color paintings. Now Little B is painting. To prevent him from drawing messy painting, Little D asks you to write a program to maintain following operations. The specific format of these operations is as follows.

0 : clear all the points.

1 x y c : add a point which color is c at point (x,y).

2 x y1 y2 : count how many different colors in the square (1,y1) and (x,y2). That is to say, if there is a point (a,b) colored c, that 1≤a≤x and y1≤b≤y2, then the color c should be counted.

3 : exit.

 

Input

The input contains many lines. 

Each line contains a operation. It may be '0', '1 x y c' ( 1≤x,y≤106,0≤c≤50 ), '2 x y1 y2' (1≤x,y1,y2≤106 ) or '3'. 

x,y,c,y1,y2 are all integers.

Assume the last operation is 3 and it appears only once.

There are at most 150000 continuous operations of operation 1 and operation 2. 

There are at most 10 operation 0. 

 

Output

For each operation 2, output an integer means the answer .

 

Sample Input

01 1000000 1000000 501 1000000 999999 01 1000000 999999 01 1000000 1000000 492 1000000 1000000 10000002 1000000 1 100000001 1 1 12 1 1 21 1 2 22 1 1 21 2 2 22 1 1 21 2 1 32 2 1 22 10 1 22 10 2 201 1 1 12 1 1 11 1 2 12 1 1 21 2 2 12 1 1 21 2 1 12 2 1 22 10 1 22 10 2 23

 

Sample Output

23122331111111

 

Source

2017ACM/ICPC广西邀请赛-重现赛（感谢广西大学）

解题思路：刚开始觉得这题很麻烦，因为可能会有点覆盖的情况，就是一个点先被一种颜色覆盖，然后被另外一种颜色覆盖，那么之前的颜色就没有了，但是，这题不会出现这种情况，就算被覆盖了，也算有两种颜色，从样例就可以看出，那么问题就变得很简单了，我们用50棵线段树搞一搞就行，我们按y坐标建树，维护x的最小值（为什么这样做，因为每次询问的矩形的左下点的横坐标是1，所以这样做可以化双边界约束为单边界，我们维护最小值就行），这里50棵线段树可能会超内存，所以要边更新，边建树，这样动态建树的话空间复杂度为O(n * logn)，因为每次更新，最多消耗logn个顶点。

#include <bits/stdc++.h>using namespace std;const int maxn = 1000000 + 10;int inf = 0x3f3f3f3f;struct node{    int l, r;    int lson, rson;    int sum;    int Min;    node()    {        sum = 0;        Min = inf;        lson = rson = 0;    }}Node[maxn<<2];int tot;int root[55];int judge;void update(int &rt, int l, int r, int value, int Mi){    if(!rt) rt = ++tot;    Node[rt].l = l;    Node[rt].r = r;    Node[rt].sum++;    Node[rt].Min = min(Node[rt].Min, Mi);    if(l == r) return;    int mid = (l + r)>>1;    if(value <= mid) update(Node[rt].lson, l, mid, value, Mi);    else update(Node[rt].rson, mid + 1, r, value, Mi);}void query(int rt, int l, int r, int up){    if(!rt || judge) return;    if(Node[rt].l == l && Node[rt].r == r)    {        if(Node[rt].Min <= up && Node[rt].sum > 0)        {            judge = 1;            return;        }        return;    }    int mid = (Node[rt].l + Node[rt].r)>>1;    if(r <= mid) query(Node[rt].lson, l, r, up);    else if(l >= mid + 1) query(Node[rt].rson, l, r, up);    else    {        query(Node[rt].lson, l, mid, up);        query(Node[rt].rson, mid + 1, r, up);    }}void init(){    tot = 0;    memset(root, 0, sizeof(root));    for(int i = 0; i < (maxn<<2); i++)    {        Node[i].lson = Node[i].rson = Node[i].sum = 0;        Node[i].Min = inf;    }}int main(){    int op;    init();    while(~scanf("%d", &op) && op != 3)    {        if(op == 0) init();        else if(op == 1)        {            int x, y, c;            scanf("%d%d%d", &x, &y, &c);            update(root[c], 1 , 1000000, y, x);        }        else if(op == 2)        {            int x, y1, y2;            scanf("%d%d%d", &x, &y1, &y2);            int ans = 0;            for(int i = 0; i <= 50; i++)            {                judge = 0;                query(root[i], y1, y2, x);                if(judge)                {                    ans++;                }            }            printf("%d\n", ans);        }    }    return 0;}