Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

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Problem G: 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.

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 r_i 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 ≤ r_i ≤ 10000000. After the i-th poster is placed, it entirely covers all wall segments numbered l_i, l_i+1 ,... , r_i.

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

Output for sample input

4

---

Author: Adapted from VI AMPwPZ by P. Rudnicki

題目描述：

給定每個海報張貼的區間，按照順序貼上去，求最後能看到幾種海報。

如果能看到同一張海報，卻被分開到好幾個地方，只能算看到一種。

題目解法：

需要離散化，離散化時特別小心，需要多離散化一些點，否則會有下方給的測資錯誤。

#include <stdio.h>
#include <map>
#include <string.h>
using namespace std;
int Tree[1048576];
int query(int k, int l, int r, int ql, int qr) {
    if(l > r)    return 1;
    if(ql <= l && r <= qr)
        return Tree[k];
    if(Tree[k])    return 1;
    int m = (l + r)/2;
    if(m >= qr)
        return query(k<<1, l, m, ql, qr);
    if(m < ql)
        return query(k<<1|1, m+1, r, ql, qr);
    return query(k<<1, l, m, ql, qr) &&
                query(k<<1|1, m+1, r, ql, qr);
}
void modify(int k, int l, int r, int ql, int qr) {
    if(l > r) {
        Tree[k] = 1;
        return;
    }
    if(ql <= l && r <= qr) {
        Tree[k] = 1;
        return ;
    }
    if(Tree[k])    return ;
    int m = (l + r)/2;
    if(m >= qr)
        modify(k<<1, l, m, ql, qr);
    else if(m < ql)
        modify(k<<1|1, m+1, r, ql, qr);
    else {
        modify(k<<1, l, m, ql, qr);
        modify(k<<1|1, m+1, r, ql, qr);
    }
    Tree[k] = Tree[k<<1]&Tree[k<<1|1];
}
int main() {
    int testcase, n;
    int i, j, k, L[10005], R[10005];
    scanf("%d", &testcase);
    while(testcase--) {
        scanf("%d", &n);
        map<int, int> RR;
        for(i = 0; i < n; i++) {
            scanf("%d %d", &L[i], &R[i]);
            RR[L[i]] = 1;
            RR[L[i]-1] = 1;
            RR[L[i]+1] = 1;
            RR[R[i]] = 1;
            RR[R[i]-1] = 1;
            RR[R[i]+1] = 1;
        }
        int size = 1;
        for(map<int, int>::iterator it = RR.begin();
            it != RR.end(); it++)
            it->second = size++;
        memset(Tree, 0, sizeof(Tree));
        int ret = 0;
        for(i = n-1; i >= 0; i--) {
            int l = RR[L[i]], r = RR[R[i]];
            // printf("[%d, %d]\n", l, r);
            int v = query(1, 1, size, l, r);
            if(v == 0)
                ret++;
            modify(1, 1, size, l, r);
        }
        printf("%d\n", ret);
    }
    return 0;
}
/*
1
3
1 10
1 3
6 10

*/