題目連結：

題意概括：

依次貼上 n 張海報，每張海報會覆蓋一個區間。後貼上的海報會覆蓋前貼的海報，問最後可以看見幾張海報

這裡的區間不是拿兩端點來維護的，是直接按單位最小區間來編號。如 [3, 5] 區間是由編號為 3、4、5的區間組成的

資料範圍：

題解分析：

其實就是一個區間塗色問題，能看見幾張海報問的就是有幾種顏色，資料結構用線段樹

線段樹維護的值是區間的顏色：

• 如果區間內的所有線段顏色相同，則該區間的顏色就是這個顏色
• 反之，該區間的顏色編號就是 INF，表示區間顏色不一致

但是，不能直接建樹。因為區間的長度上限達 10000000，就算是 O(nlogn) 也會TLE。還可能MLE

由於海報的數量上限為 1e4 ，我們可以用離散化方法預處理資料，把區間的長度壓縮壓縮到 1e4

關於離散化的方法：

注意的點：

• 這題的資料有問題，n 明顯大於 10000，我之前一直RE，把陣列開到 4e4 才過的
• 當 n == 0 時要特判
• 處理好離散化的bug，用加點的方法

AC程式碼：

#include <stdio.h>
#include <memory.h>
#include <algorithm>
using namespace std;
const int MAXN = 4e4 + 10;
const int INF = 0x3f3f3f3f;
int origin[MAXN], tree[MAXN<<2], lazy[MAXN<<2], ans[MAXN], last;
int disc[MAXN], L[MAXN], R[MAXN];          //離散陣列，左端點，右端點

void pushup(int p) {
    if (tree[p << 1] == tree[p << 1 | 1])
        tree[p] = tree[p << 1];
    else
        tree[p] = INF;
}

void pushdown(int p, int l, int r) {
    if (lazy[p] != INF) {
        lazy[p << 1] = lazy[p];
        lazy[p << 1 | 1] = lazy[p];
        tree[p << 1] = lazy[p];  //左右兒子結點均按照需要加的值總和更新結點資訊
        tree[p << 1 | 1] = lazy[p];
        lazy[p] = INF;
    }
}

void build(int p, int l, int r) {
    if (l == r) {
        tree[p] = origin[l];
        return;
    }
    int mid = (l + r) >> 1;
    build(p << 1, l, mid);
    build(p << 1 | 1, mid + 1, r);
    pushup(p);
}

void update(int p, int l, int r, int ql, int qr, int v) {   //區間更新
    if (ql <= l && r <= qr) {       //當前區間在更新區間內
        tree[p] = v;
        lazy[p] = v;
        return;
    }
    int mid = (l + r) >> 1;
    pushdown(p, l, r);
    if (ql <= mid)  update(p << 1, l, mid, ql, qr, v);
    if (qr > mid)  update(p << 1 | 1, mid + 1, r, ql, qr, v);
    pushup(p);
}


void query(int p, int l, int r, int ql, int qr) {

    if (tree[p] != INF) {
        if (tree[p] != last)
            ans[tree[p]] ++;
        last = tree[p];
        return;
    }
    if (l == r) {last = INF; return;}
    int mid = (l + r) >> 1;
    pushdown(p, l, r);
    query(p << 1, l, mid, ql, qr);
    query(p << 1 | 1, mid + 1, r, ql, qr);
}

int main () {

    int n, T;
    scanf("%d", &T);
    while(T--) {
        int tot = 0;
        scanf("%d", &n);
        if (n) {
        for (int k = 0; k < n; k++) {
            scanf("%d%d", L + k, R + k);
            disc[tot++] = L[k];
            disc[tot++] = R[k];
        }
        //離散化
        sort(disc, disc + tot);
        int size = (int)(unique(disc, disc + tot) - disc);
        int temp = size;
        for (int i = 1; i < temp; i++)
            if (disc[i] - disc[i - 1] > 1)
                disc[size++] = disc[i - 1] + 1;
        
        sort(disc, disc + size);
        for (int i = 0; i < n; i ++) {
            L[i] = (int)(lower_bound(disc, disc + size, L[i]) - disc + 1);
            R[i] = (int)(lower_bound(disc, disc + size, R[i]) - disc + 1);
        }
        
        memset(lazy, INF, sizeof(lazy));
        memset(origin, INF, sizeof(origin));
        memset(ans, 0, sizeof(ans));

        last = INF;
        build(1, 1, size);
        
        for (int i = 0; i < n; i ++)
            update(1, 1, size, L[i], R[i], i + 1);
            
        query(1, 1, size, 1, size);
        int cnt = 0;
        for (int i = 1; i <= size; i ++)
            if (ans[i]) cnt++;
        printf("%d\n", cnt);
        }
        else printf("0\n");
    }
}
 

                                                  Mayor's posters

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