Mayor's posters

Mayor's posters Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 2528

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

由于数据太大，而中间数据大多没用，我们只需要判断从前往后贴的过程之后，人们可以看出来几张不同海报（每张都不同，即看是否海报被完全遮住，当然在贴海报的过程中，只有后边可以遮住前边的，
所以我们从前往后判断，是否当前海报被遮住，如果没有，肯定也不会被比他贴的还早的海报遮住，就是可以看到，就加一。
由于数据太大，太多，太多useless，so我们用线段树+离散化进行处理

#include<iostream>
#include<cstdio>
#include<algorithm>

using namespace std;

#define N 210000  // 由于数据太多

struct node
{
    int l, r;
}val[N];

struct point
{
    int l, r, flag;
}tree[N*4];

int x[N*2], h[N*1000];  // h数组为什么存这么多，因为1 <= i <= n, 1 <= l i <= ri <= 10000000.

void Build(int root, int l, int r)  // 建树
{
    tree[root].l = l;
    tree[root].r = r;

    tree[root].flag = false;

    if(l == r)
        return ;

    Build(root*2, l, (l+r)/2);
    Build(root*2+1, (l+r)/2+1, r);

}

void update(int root)  // 更新
{
    if(tree[root*2].flag && tree[root*2+1].flag)
        tree[root].flag = 1;
}

int query(int root, int l, int r)  // 看是否可以看到，如果可以看到return 1；else return0；
{
    if(tree[root].flag)  // 如果被遮住，return0；
        return 0; 

    if(tree[root].l == tree[root].r)  // 如果本节点没有被遮住，而且到了叶节点，就返回1，本节点flag置为1，即已被遮住
    { 
        tree[root].flag = 1;
        return 1;
    }

    update(root);  // 更新root节点

    int mid = (tree[root].l+tree[root].r)/2;

    if(r <= mid)
        return query(2*root, l, r);
    else if(l > mid)
        return query(2*root+1, l, r);
    else
        return query(2*root, l, mid)+query(2*root+1, mid+1, r);

}
int main()
{
    int c, n;

    scanf("%d", &c);

    while(c--)
    {
        scanf("%d", &n);
        int k = 0;

        for(int i = 1; i <= n; i++)
        {
            scanf("%d%d", &val[i].l, &val[i].r);
            x[k++] = val[i].l, x[k++] = val[i].r;  // x数组存各个海报的位置
        }

        sort(x, x+k);
        k = unique(x, x+k)-x;  // 去重，排序

        Build(1, 0, k-1); // 建树，总共有k-1个点，就有k-1个节点

        for(int i = 0; i < k; i++)
            h[x[i]] = i;  // 把各个节点的值离散化，就是把中间的有的没得全没了，只留海报左右两个端点的值存到h数组，作为节点的l，r值
 
        int ans = 0;
        for(int i = n; i > 0; i--)
            if(query(1, h[val[i].l], h[val[i].r]))  // 查询
                ans++;
                
        printf("%d\n", ans);
    }
    return 0;
}

 

转载于:https://www.cnblogs.com/Tinamei/p/4694559.html