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。

49愛的鼓勵 6訂閱站台

The skyline of Singapore as viewed from the Marina Promenade (shown on the left) is one of the iconic scenes of Singapore. Country X would also like to create an iconic skyline, and it has put up a call for proposals. Each submitted proposal is a description of a proposed skyline and one of the metrics that country X will use to evaluate a proposed skyline is the amount of overlap in the proposed sky-line.

As the assistant to the chair of the skyline evaluation committee, you have been tasked with determining the amount of overlap in each proposal. Each proposal is a sequence of buildings, b₁, b₂,..., b_n , where a building is specified by its left and right endpoint and its height. The buildings are specified in back to front order, in other words a building which appears later in the sequence appears in front of a building which appears earlier in the sequence.

The skyline formed by the first k buildings is the union of the rectangles of the first k buildings (see Figure 4). The overlap of a building, b_i , is defined as the total horizontal length of the parts of b_i , whose height is greater than or equal to the skyline behind it. This is equivalent to the total horizontal length of parts of the skyline behind b_i which has a height that is less than or equal to h_i , where h_i is the height of building b_i . You may assume that initially the skyline has height zero everywhere.

Input 

The input consists of a line containing the number c of datasets, followed by c datasets, followed by a line containing the number `0'.

The first line of each dataset consists of a single positive integer, n (0 < n < 100000) , which is the number of buildings in the proposal. The following n lines of each dataset each contains a description of a single building. The i -th line is a description of building b_i . Each building b_i is described by three positive integers, separated by spaces, namely, l_i , r_i and h_i , where l_i and r_j (0 < l_i < r_i <=100000) represents the left and right end point of the building and h_i represents the height of the building.

Output 

The output consists of one line for each dataset. The c -th line contains one single integer, representing the amount of overlap in the proposal for dataset c . You may assume that the amount of overlap for each dataset is at most 2000000.

Note: In this test case, the overlap of building b₁ , b₂ and b₃ are 6, 4 and 4 respectively. Figure 4 shows how to compute the overlap of building b₃ . The grey area represents the skyline formed by b₁ and b₂ and the black rectangle represents b₃ . As shown in the figure, the length of the skyline covered by b₃ is from position 3 to position 5 and from position 11 to position 13, therefore the overlap of b₃ is 4.

Sample Input 

1 
3 
5 11 3 
1 10 1 
3 13 2 
0

Sample Output 

14

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先說說題目, 建築物從海岸線開始退, 每個建築物可以當作一個長方形平行海岸,
而每個建築物都有其高度, 如果被前面的建築物遮蔽的區塊, 則在這建築物中的一小段區間
是無法看到海岸的, 計算所有建築物可以看到海岸的區間長度總和。

雖然說是線段樹, 但操作並不是 O(logn), 跟輸入有關, 不過能快就快了,
記錄每個區間的最大高度與最小高度, 最大高度用來判定能不能計算這個區間可以被看到,
而最小高度則用來判定這個區間有沒有機會被這個建築物的區間看到海岸,

因此最後會簡單地想到, 一個線段樹的區間如果被 matched, 不一定就這麼結束,
會直到找到可以更新的區間並加以計算。

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
struct Node {
    int mx, mn, label;
};
Node ST[(1<<18)+1];
long long ret;
void update(int k) {
    ST[k].mx = max(ST[k<<1].mx, ST[k<<1|1].mx);
    ST[k].mn = min(ST[k<<1].mn, ST[k<<1|1].mn);
}
void downdate(int k) {
    if(ST[k].label) {
        ST[k<<1].mx = ST[k<<1].mn = ST[k<<1].label = ST[k].label;
        ST[k<<1|1].mx = ST[k<<1|1].mn = ST[k<<1|1].label = ST[k].label;
        ST[k].label = 0;
    }
}
void modify(int k, int l, int r, int &ql, int &qr, int h) {
    if(l > r)
        return;
    if(l != r)
        downdate(k);
    if(ST[k].mn > h)
        return;
    if(ql <= l && r <= qr) {
        if(h >= ST[k].mx) {
            ret += r-l+1;
            ST[k].mx = ST[k].mn = ST[k].label = h;
            return;
        }
        // still update sub-interval
    }
    int m = (l+r)>>1;
    if(ql > m)
        modify(k<<1|1, m+1, r, ql, qr, h);
    else if(qr <= m)
        modify(k<<1, l, m, ql, qr, h);
    else {
        modify(k<<1, l, m, ql, qr, h);
        modify(k<<1|1, m+1, r, ql, qr, h);
    }
    update(k);
}
int main() {
    int testcase, n, x, y, h;
    int i, j;
    scanf("%d", &testcase);
    while(testcase--) {
        scanf("%d", &n);
        memset(ST, 0, sizeof(ST));
        ret = 0;
        for(i = 0; i < n; i++) {
            scanf("%d %d %d", &x, &y, &h);
            y--;
            modify(1, 1, 100000, x, y, h);
        }
        printf("%lld\n", ret);
    }
    return 0;
}