poj 2528

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

树状数组的应用。离散化的其中一种方法就是用一个数组存取原来的数，然后排序，然后用lower_bound 来找排序后的下标。然后完成离散化。

#include <cstdio>#include <cstring>#include <algorithm>#define LL rt<<1#define RR rt<<1|1#define lson l,m,LL#define rson m+1,r,RRconst int maxn=11111;bool hash1[10010];  //仅仅是用来记录这个点是否输出了避免重复点。int col[maxn<<4]; //存的是区间结点，维护的是画的编号，区间结点维护画的编号using namespace std;//线段树的应用住要集中在边界身上，即每一个节点代表一个区间。要维护的东西放在结点上，用数组或者结构体都可以，每次比较边界//用二分法查找边界（求中点，用宏会更方便struct node{    int l,r; //用一个结构体来保存边界，保存的是油画的左右边界}q[10010];int ans;int x[maxn<<2];void pushdown(int rt){    if(col[rt]!=-1)    {        col[LL]=col[RR]=col[rt];        col[rt]=-1;    }}void update(int L,int R,int c,int l,int r,int rt){    if(L<=l&&r<=R)    {        col[rt]=c;        return ;    }    pushdown(rt);     //往下更新子区间结点的画编号    int m=(l+r)>>1;    if(L<=m) update(L,R,c,lson);    if(R>m) update(L,R,c,rson);    }void query(int l,int r,int rt){    if(col[rt]!=-1)    {        if(!hash1[col[rt]]) ans++;        hash1[col[rt]]=true;        return ;    }    if(l==r) return;    int m=(l+r)>>1;    query(lson);    query(rson);}int main(){    int n,i,j,t;    scanf("%d",&t);    while(t--)    {        int cnt=0;        scanf("%d",&n);        for(i=0;i<n;i++)        {            scanf("%d%d",&q[i].l,&q[i].r);            x[cnt++]=q[i].l;x[cnt++]=q[i].r; //离散化 把边界都都进去        }        sort(x,x+cnt);        int m=1;        for(i=1;i<cnt;i++)            if(x[i]!=x[i-1])    //保证边界的唯一性，去重                x[m++]=x[i];        for(i=m-1;i>=1;i--)            if(x[i]!=x[i-1]+1)                x[m++]=x[i-1]+1;  //在分隔区间大于1的的区间加点，让子区区间变多，不然的话每幅画只有两个点，很容易被误判成覆盖。        sort(x,x+m);        memset(col,-1,sizeof(col));        for(i=0;i<n;i++)        {            int l=lower_bound(x,x+m,q[i].l)-x; //离散化后这样找到左边界的下标 其实还是找左边界的下表（相当于找左边界）            int r=lower_bound(x,x+m,q[i].r)-x;// 离散化后这样找右边界的下标            update(l,r,i,0,m,1);       //令在油画边界里的子区间全都更新为油画的编号        }        memset(hash1,false,sizeof(hash1));        ans=0;        query(0,m,1);        printf("%d\n", ans);    }}