POJ 2528 Mayor's posters 线段树

/**
* @file main.cpp
* @brief 离散化+线段树
*        1.离散化,源数据的范围是1-10000000,直接使用线段树求解会超内存,也不优雅
*          数据个数只有10000,离散化后只有20000个点
*        2.线段树求解,我感觉的线段树的精髓在于插入/染色的时候把父节点的值传递给子节点这个操作,在代码中已经特别注释出来
*
* @author yekeren
* @version 1.0.0
* @date 2013-06-09
*/
#include <stdio.h>
#include <set>
#include <map>
#include <vector>

#define LEFT(x) (((x) << 1) + 1)
#define RIGHT(x) (((x) << 1) + 2)

unsigned short k[100000] = { 0 };
std::set<unsigned short> counter;

/**
* @brief 线段树插入操作
* @param root
* @param left
* @param right
* @param a
* @param b
* @param c
*/
void insert(int root, int left, int right, int a, int b, int c)
{
if (a <= left && b >= right)
{
k[root] = c;
return;
}

if (k[root])    ///<精髓
{
k[LEFT(root)]  = k[root];
k[RIGHT(root)] = k[root];
k[root] = 0;
}

int mid = (left + right) >> 1;
if (b <= mid) {
insert(LEFT(root), left, mid, a, b, c);
} else if (a > mid) {
insert(RIGHT(root), mid + 1, right, a, b, c);
} else {
insert(LEFT(root), left, mid, a, mid, c);
insert(RIGHT(root), mid + 1, right, mid + 1, b, c);
}
}

/**
* @brief 求最终覆盖结果
* @param root
* @param left
* @param right
*/
void sum(int root, int left, int right)
{
if (k[root]) {
counter.insert(k[root]);
return;
}

if (left == right) {
return;
}

int mid = (left + right) >> 1;
sum(LEFT(root), left, mid);
sum(RIGHT(root), mid + 1, right);
}

int main(int argc, char *argv[])
{
int c, n;
scanf("%d", &c);

while (c--)
{
scanf("%d", &n);

int data[10010][2] = { 0 };
std::map<int, int> mapdict;

///<离散化
for (int i = 0; i < n; ++i)
{
scanf("%d%d", &data[i][0], &data[i][1]);
mapdict.insert(std::pair<int, int>(data[i][0], 0));
mapdict.insert(std::pair<int, int>(data[i][1], 0));
}

int m = 0;
std::map<int, int>::iterator iter;
for (iter = mapdict.begin(); iter != mapdict.end(); ++iter) {
iter->second = ++m;
}

///<线段树插入
for (int i = 0; i < n; ++i)
{
int a = mapdict[data[i][0]];
int b = mapdict[data[i][1]];

insert(0, 1, m, a, b, 1 + i);
}

///<输出最终覆盖结果
counter.clear();
sum(0, 1, m);
printf("%d\n", counter.size());
}
return 0;
}