POJ 2528 Mayor's posters(线段树 + 离散化--静态实现)

Mayor's posters

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 58066		Accepted: 16794

Description

The citizens of Bytetown, AB, could not stand that the candidates in the mayoral election campaign have been placing their electoral posters at all places at their whim. The city council has finally decided to build an electoral wall for placing the posters
and introduce the following rules: 
Every candidate can place exactly one poster on the wall. 

All posters are of the same height equal to the height of the wall; the width of a poster can be any integer number of bytes (byte is the unit of length in Bytetown). 

The wall is divided into segments and the width of each segment is one byte. 

Each poster must completely cover a contiguous number of wall segments.

They have built a wall 10000000 bytes long (such that there is enough place for all candidates). When the electoral campaign was restarted, the candidates were placing their posters on the wall and their posters differed widely in width. Moreover, the candidates
started placing their posters on wall segments already occupied by other posters. Everyone in Bytetown was curious whose posters will be visible (entirely or in part) on the last day before elections. 

Your task is to find the number of visible posters when all the posters are placed given the information about posters' size, their place and order of placement on the electoral wall. 

Input

The first line of input contains a number c giving the number of cases that follow. The first line of data for a single case contains number 1 <= n <= 10000. The subsequent n lines describe the posters in the order in which they were placed. The i-th line among
the n lines contains two integer numbers li and ri which are the number of the wall segment occupied by the left end and the right end of the i-th poster, respectively. We know that for each 1 <= i <= n, 1 <= li <= ri <= 10000000. After
the i-th poster is placed, it entirely covers all wall segments numbered li, li+1 ,... , ri.
Output

For each input data set print the number of visible posters after all the posters are placed. 

The picture below illustrates the case of the sample input. 



Sample Input

1
5
1 4
2 6
8 10
3 4
7 10

Sample Output

4

Source

Alberta Collegiate Programming Contest 2003.10.18

之前写过线段树的动态实现，现在借此题搞搞静态的实现方法

静态实现不借助指针指向两个子节点，而是一个线性结构，借助计算（node*2）(node*2 - 1)来指明两个子节点

不过这题难点是离散化和判断海报能否最终被看到

1、这题的离散化是有点特殊的

例如这样一个样例

1 10

1 5

6 10

那么这个样例应该答案是2

1 5 6 10，离散化后是0 1 2 3，答案是对的

如果样例换成

1 10

1 4
6 10

1 4 6 10，离散化后还是0 1 2 3，这样答案就错了，因为没有考虑到4和6之间还有空

我们只要加一层循环就可以了，看一下如果后一个数字a[i]比前一个a[i-1]大的超过1，那么就在原有数组的基础上加上一个a[i-1]+1,那么1 4 6 10离散化后就是1 2 4 5 6 7 10 

2、在判断海报是，从后往前判断，这样省事很多

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAXN = 1e5 + 100;

struct Post {
int l, r;
} post[MAXN];

struct Node {
int st, ed;
int val;
bool vis;
} tree[MAXN << 4];

int t, n;
int disp[MAXN << 4];
bool canPost;

void Build(int node, int st, int ed) {
tree[node].st = st;
tree[node].ed = ed;
tree[node].val = 0;
tree[node].vis = false;
if (st == ed) {
return;
}
int mid = st + ed >> 1;
Build(node << 1, st, mid);
Build(node << 1 | 1, mid + 1, ed);
}

void PushUp(int node) {
if (tree[node << 1].vis && tree[node << 1 | 1].vis) {
tree[node].vis = true;
}
}

void Update(int node, int st, int ed, int val) {
if (tree[node].vis) {
return;
}
if (tree[node].st == st && tree[node].ed == ed) {
tree[node].vis = true;
canPost = true;
return;
}
if (tree[node << 1].ed >= ed) {
Update(node << 1, st, ed, val);
}
else if (tree[node << 1 | 1].st <= st) {
Update(node << 1 | 1, st, ed, val);
}
else {
Update(node << 1, st, tree[node << 1].ed, val);
Update(node << 1 | 1, tree[node << 1 | 1].st, ed, val);
}
PushUp(node);
}

int main() {
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
#endif
scanf("%d", &t);
while (t--) {
scanf("%d", &n);
int cnt = 0;
for (int i = 0; i < n; ++i) {
scanf("%d%d", &post[i].l, &post[i].r);
disp[cnt++] = post[i].l;
disp[cnt++] = post[i].r;
}
sort(disp, disp + cnt);
cnt = unique(disp, disp + cnt) - disp;
for (int i = 1, ed = cnt; i < ed; ++i) {
if (disp[i] - disp[i - 1] != 1) {
disp[cnt++] = disp[i - 1] + 1;
}
}

sort(disp, disp + cnt);
Build(1, 1, cnt);
int ans = 0;

for (int i = n - 1; i >= 0; --i) {
int hashl = lower_bound(disp, disp + cnt, post[i].l) - disp + 1;
int hashr = lower_bound(disp, disp + cnt, post[i].r) - disp + 1;
canPost = false;
Update(1, hashl, hashr, i + 1);
if (canPost) ++ans;
}
printf("%d\n", ans);
}
return 0;
}