[UVA][段] 10587 - Mayor's posters＠Morris' Blog｜PChome Online 人新台

Morris' Blog 我在摸索我存在的意、生命的意、涵及值。有真正神手般的害，套用模式是大部分的情。那害的我存在的意何，托出神的存在？有候不能怪人不我，我已始省思，什我有人的原因，而是全要靠人。有候我我不同，那是因我做不到很多人能到的事情。我不到、玩不起、得不明、理解得不多，我缺少的文能力，就是一切一切能的判定。此，我不得能做些人有助的事情。我知道我自己是不服的性格，但在我不得不。我有能力，因此我必。我到底世界作什？充不定性，而我是法出拔萃，想一般人似乎也回不了了。越大的挑需要越力的支持，而我的支持是什？最近人，但都是失的局，人能接受，不需要言，不需要通，一切所想都表不出。
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2014-02-17 09:22:27| 人2,094| 回0 | 上一篇 | 下一篇

[UVA][段] 10587 - Mayor's posters

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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

Author: Adapted from VI AMPwPZ by P. Rudnicki

目描述：

定每海的，按照序上去，求最後能看到海。

如果能看到同一海，被分到好地方，只能算看到一。

目解法：

需要散化，散化特小心，需要多散化一些，否有下方的。

#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);
  &nsp; 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

*/