poj2528 Mayor's posters(线段树区间修改+特殊离散化）

最新推荐文章于 2020-03-10 14:07:43 发布

转载最新推荐文章于 2020-03-10 14:07:43 发布 · 52 阅读

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原文链接：http://www.cnblogs.com/TnT2333333/p/6023903.html

本文介绍了一种使用线段树解决特定区间覆盖问题的方法，旨在确定在一系列有序区间覆盖后，有多少个不同的区间仍可见。

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.

题意：给你一系列的区间有顺序的，按顺序将给定区间染上各不相同的颜色，最后问你一共能看到几种颜色。

我的想法是倒过来考虑，因为在最后涂的颜色不会被覆盖掉。染完色就讲这个区间全部赋值为1，到下个区间如果这个区间内所有值都为1，

那么这个颜色就看不到了，被覆盖掉了。实现这种方法可以借助线段树区间更新，接下来就是考虑怎么建树了，由于数的范围较大，但给的

区间比较小，所以可以离散化一下。但这题的离散化有些特殊，不能普通的离散化。举一个例子给你3个区间

（1，10）（1，6）（8，10）正常离散化后是（1，4）（1，2）（3，4）结果是2但是正确答案是1！，如何解决这个问题呢？可以将两个

相差大于1的数离散化时在他与下一个之间插入一个值，如给的例子离散化后的结果（1，7）（1，3）（5，7）及将1，6，8，10离散化为

1，（2），3，（4），5，（6），7（括号中的数为插入的值。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
const int M = 2e5 + 10;
int a[M] , b[M] , c[2 * M] , d[2 * M] , e[4 * M];
struct TnT {
    int l , r , num , add;
}T[M << 2];
int re;
void push(int p) {
    if(T[p].add) {
        T[p << 1].num = (T[p << 1].r - T[p << 1].l + 1);
        T[(p << 1) | 1].num = (T[(p << 1) | 1].r - T[(p << 1) | 1].l + 1);
        T[p << 1].add = T[p].add;
        T[(p << 1) | 1].add = T[p].add;
        T[p].add = 0;
    }
}
void build(int l , int r , int p) {
    int mid = (l + r) >> 1;
    T[p].l = l , T[p].r = r , T[p].num = 0 , T[p].add = 0;
    if(T[p].l == T[p].r) {
        return ;
    }
    build(l , mid , p << 1);
    build(mid + 1 , r , (p << 1) | 1);
    T[p].num = T[p << 1].num + T[(p << 1) | 1].num;
}
void updata(int l , int r , int p) {
    int mid = (T[p].l + T[p].r) >> 1;
    if(T[p].l == l && T[p].r == r) {
        T[p].add = 1;
        T[p].num = (r - l + 1);
        return ;
    }
    push(p);
    if(mid < l) {
        updata(l , r , (p << 1) | 1);
    }
    else if(mid >= r) {
        updata(l , r , p << 1);
    }
    else {
        updata(l , mid , p << 1);
        updata(mid + 1 , r , (p << 1) | 1);
    }
    T[p].num = T[p << 1].num + T[(p << 1) | 1].num;
}
int query(int l , int r , int p) {
    int mid = (T[p].l + T[p].r) >> 1;
    if(T[p].l == l && T[p].r == r) {
        return T[p].num;
    }
    push(p);
    T[p].num = T[p << 1].num + T[(p << 1) | 1].num;
    if(mid < l) {
        return query(l , r , (p << 1) | 1);
    }
    else if(mid >= r) {
        return query(l , r , p << 1);
    }
    else {
        return query(l , mid , p << 1) + query(mid + 1 , r , (p << 1) | 1);
    }
}
int search(int ll, int hh, int xx) {
    int mm;
    while (ll <= hh) {
        mm = (ll + hh) >> 1;
        if (e[mm] == xx) return mm;
        else if (e[mm] > xx)  hh = mm - 1;
        else ll = mm + 1;
    }
    return -1;
}
int main()
{
    int t;
    scanf("%d" , &t);
    while(t--) {
        int n;
        scanf("%d" , &n);
        int gg = 0;
        for(int i = 1 ; i <= n ; i++) {
            scanf("%d%d" , &a[i] , &b[i]);
            c[++gg] = a[i];
            c[++gg] = b[i];
        }
        sort(c + 1 , c + gg + 1);
        int mm = 0;
        c[gg + 1] = -1;
        for(int i = 1 ; i <= gg ; i++) {
            if(c[i] != c[i + 1]) {
                d[++mm] = c[i];
            }
        }
        e[1] = d[1];
        int mt = 1;
        for(int i = 2 ; i <= mm ; i++) {
            if(d[i] - d[i - 1] > 1) {
                e[++mt] = d[i - 1] + 1;
                e[++mt] = d[i];
            }
            else {
                e[++mt] = d[i];
            }
        }
        //        for(int i = 1 ; i <= mt ; i++) {
        //            cout << e[i] << ' ';
        //        }
        build(1 , mt + 1 , 1);
        int count = 0;
        for(int i = n ; i >= 1 ; i--) {
            int r = search(1 , mt , b[i]);
            int l = search(1 , mt , a[i]);
            re = query(l , r ,  1);
            //cout << re << endl;
            if(re < r - l + 1) {
                count++;
            }
            updata(l , r , 1);
        }
        printf("%d\n" , count);
    }
    return 0;
}

转载于:https://www.cnblogs.com/TnT2333333/p/6023903.html