HDU6183-Color it Time

Color it

                                                                    Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 132768/132768 K (Java/Others)
                                                                                                  Total Submission(s): 579    Accepted Submission(s): 153

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）

 

有四种操作：
0:清除所有点
1 x y c : 给点(x, y)添加一种颜色c（颜色不会覆盖）
2 x y1 y2 : 在(0, y1)与(x, y2)所围成的矩形里有多少种颜色
3 : 程序结束

解题思路：颜色最多51种。我们就建51棵线段树。 每个线段树按y轴建树，每个结点的值是在范围内的最小的x值

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <map>#include <cmath>#include <set>#include <stack>#include <queue>#include <vector>#include <bitset>#include <functional>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;int s[55], L[1000009], R[1000009], v[1000009];int tot, x, y, w, yy1, yy2, flag;void update(int &k, int l, int r, int y, int x){if (!k){L[k = tot] = 0, R[tot++] = 0;v[k] = x;}v[k] = min(v[k], x);if (l == r) return;int mid = (l + r) >> 1;if (y <= mid) update(L[k], l, mid, y, x);else update(R[k], mid + 1, r, y, x);}void query(int k, int l, int r, int ll, int rr){if (flag || !k) return;if (ll <= l && rr >= r){if (v[k] <= x) flag = 1;return;}int mid = (l + r) >> 1;if (ll <= mid) query(L[k], l, mid, ll, rr);if (rr > mid) query(R[k], mid + 1, r, ll, rr);}int main(){int p;while (~scanf("%d", &p)){if (p == 3) break;if (p == 0){memset(s, 0, sizeof s);tot = 1;}else if (p == 1){scanf("%d%d%d", &x, &y, &w);update(s[w], 1, 1000000, y, x);}else if (p == 2){scanf("%d%d%d", &x, &yy1, &yy2);int ans = 0;for (int i = 0; i <= 50; i++){flag = 0;query(s[i], 1, 1000000, yy1, yy2);if (flag) ans++;}printf("%d\n", ans);}}return 0;}