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 l
i 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 <= l
i <= ri <= 10000000. After the i-th poster is placed, it entirely covers all wall segments numbered l
i, l i+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
题意：铺壁纸，求最后能让人看到的有几个
思路：数据很大，如果单纯以距离作为下表的话会超内存，思路：我们将我们需要的点离散化，但是普通的离散化又会出现错误：比如区间[1, 10], [1, 4] , [5 ,10] 以及例子：[1, 10], [1 ,4] , [6, 10]，假如我们已经离散化了，那我们插入的区间就是: [1, 3], [1, 2], [2 ,3 ]这种插法对第一个例子是适用的，但是对第二个就不行了，解决的方法是：如果两个数的位置大于1
的话，我们就多加一个点来区别这两个点不是连在一起的
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
#define lson(x) ((x) << 1)
#define rson(x) ((x) << 1 | 1)
using namespace std;
const int maxn = 100005;
bool hash[maxn];
int li[maxn], ri[maxn], ans;
int x[maxn<<2];
struct seg {
int w;
int flag;
};
struct segment_tree {
seg node[maxn<<2];
void build(int l, int r, int pos) {
if (l == r) {
node[pos].w = -1;
return;
}
int m = l + r >> 1;
build(l, m, lson(pos));
build(m+1, r, rson(pos));
}
void push(int pos) {
if (node[pos].w != -1) {
node[lson(pos)].w = node[rson(pos)].w = node[pos].w;
node[pos].w = -1;
}
}
void modify(int l, int r, int pos, int x, int y, int z) {
if (x <= l && y >= r) {
node[pos].w = z;
return;
}
push(pos);
int m = l + r >> 1;
if (x <= m)
modify(l, m, lson(pos), x, y, z);
if (y > m)
modify(m+1, r, rson(pos), x, y, z);
}
void query(int l, int r, int pos) {
if (node[pos].w != -1) {
if (hash[node[pos].w] == false)
ans++;
hash[node[pos].w] = true;
return;
}
if (l == r)
return;
int m = l + r >> 1;
query(l, m, lson(pos));
query(m+1, r, rson(pos));
}
} tree;
int Bin(int key, int n, int arr[]) {
int l = 0, r = n-1;
while (l <= r) {
int m = l + r >> 1;
if (arr[m] == key)
return m;
if (arr[m] < key)
l = m + 1;
else r = m-1;
}
return -1;
}
int main() {
int t, n;
scanf("%d", &t);
while (t--) {
scanf("%d", &n);
int cnt = 0;
for (int i = 0; i < n; i++) {
scanf("%d%d", &li[i], &ri[i]);
x[cnt++] = li[i];
x[cnt++] = ri[i];
}
sort(x, x+cnt);
int m = 1;
for (int i = 1; i < cnt; i++)
if (x[i] != x[i-1])
x[m++] = x[i];
for (int i = m-1; i > 0; i--)
if (x[i] != x[i-1] + 1)
x[m++] = x[i-1] + 1;
sort(x, x+m);
tree.build(1, m, 1);
for (int i = 0; i < n; i++) {
int l = Bin(li[i], m, x) + 1;
int r = Bin(ri[i], m, x) + 1;
tree.modify(1, m, 1, l, r, i);
}
ans = 0;
memset(hash, false, sizeof(hash));
tree.query(1, m, 1);
printf("%d\n", ans);
}
return 0;
}
POJ - 2528 Mayor's posters （线段树+离散化）