poj 2528 Mayor's posters 【线段树 + 离散化】

Mayor's posters

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 50643		Accepted: 14675

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

题意：一个城市要竞选市长。竞选者能够在一块墙上贴海报为自己拉票，每一个人能够贴连续的一块区域。后来贴的能够覆盖前面的，问到最后一共能够看到多少张海报。

第一道离散化：（滚动数组优化） ORZ网上的大牛

#include <cstdio>
#include <cstring>
#include <algorithm>
#define MAXN 10000+100
using namespace std;
struct Node
{
    int x, y;
};
Node num[10100];
int color[MAXN<<4];
int rec[MAXN<<4];//离散化 存储
int Find(int val, int *a, int L, int R)//在a数组下标[L, R]范围里面 查找val值的下标
{
    int left = L, right = R;
    while(left <= right)
    {
        int mid = (left + right) >> 1;
        if(a[mid] == val)
            return mid;
        if(a[mid] < val)
            left = mid  + 1;
        else
            right = mid - 1;
    }
    return -1;
}
void PushDown(int o)
{
    if(color[o])
    {
        color[o<<1] = color[o<<1|1] = color[o];
        color[o] = 0;
    }
}
void update(int o, int l, int r, int L, int R, int v)
{
    if(L <= l && R >= r)
    {
        color[o] = v;
        return ;
    }
    PushDown(o);
    int mid = (l + r) >> 1;
    if(L <= mid)
        update(o<<1, l, mid, L, R, v);
    if(R > mid)
        update(o<<1|1, mid+1, r, L, R, v);
}
int vis[10100];//标记该海报是否出现过
int ans;//纪录数目
void query(int o, int l, int r)
{
    if(color[o])
    {
        if(!vis[color[o]])
        ans++,vis[color[o]] = true;
            return ;
    }
        //return ;
    if(l == r)
        return ;
    int mid = (l + r) >> 1;
    query(o<<1, l, mid);
    query(o<<1|1, mid+1, r);
}
int main()
{
    int t, N;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d", &N);
        memset(color, 0, sizeof(color));
        int len = 1;
        for(int i = 1; i <= N; i++)
        {
            scanf("%d%d", &num[i].x, &num[i].y);
            rec[len++] = num[i].x;
            rec[len++] = num[i].y;
        }
        sort(rec+1, rec+len);
        //离散化
        int RR = 2;
        for(int i = 2; i < len; i++)//滚动数组优化
        {
            if(rec[i] != rec[i-1])
                rec[RR++] = rec[i];
        }
        for(int i = RR-1; i > 1; i--)
        {
            if(rec[i] != rec[i-1] + 1)
                rec[RR++] = rec[i-1] + 1;
        }
        sort(rec+1, rec+RR);//不是RR+1
        for(int i = 1; i <= N; i++)
        {
            int l = Find(num[i].x, rec, 1, RR-1);
            int r = Find(num[i].y, rec, 1, RR-1);
            update(1, 1, RR-1, l, r, i);
        }
        memset(vis, false, sizeof(vis));
        ans = 0;
        query(1, 1, RR-1);
        printf("%d\n", ans);
    }
    return 0;
}