poj2528Mayor's posters【离散化线段树区间更新】

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 54376		Accepted: 15796

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

Sample Output

Source

Alberta Collegiate Programming Contest 2003.10.18

开学第一题~将近两个小时才过，暴露一堆问题==

说题意：在高度一定的展板上依次贴与展板一样高的海报，也就是后来的贴在上面，问最后能看到几个。开始特别没脑子的val值想设成当前区间上能看到的不同海报的个数，实在想不到update怎么写。标称是把val设成是当前最上面的颜色，智商啊啊啊啊。

update()的递归写法也错了好多次

其实最重要的是离散化的过程，由于展板分成整数单元，普通离散化不能表示，间隔超过一的中间加一点即可。

最最最最傻×的是没写scanf("%d",&n);!!!

/*******************
poj2528
2016.3.7
1088K	79MS	C++	2669B
*******************/
#include <cstdio>
#include <algorithm>
#include<cstring>
using namespace std;
#define lson l , m , rt << 1
#define rson m + 1 , r , rt << 1 | 1
const int maxn = 11111;
bool hash[maxn];
int li[maxn] , ri[maxn];
int X[maxn*3];
int col[maxn<<4];
int cnt;
void push_down(int rt)
{
    if(col[rt]!=-1)
    {
        col[rt<<1]=col[rt<<1|1]=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;
    }
    push_down(rt);
    int mid=(L+R)/2;
    if(l<=mid)update(l,r,c,L,mid,rt<<1);
    if(r>mid) update(l,r,c,mid+1,R,rt<<1|1);///
}
void query(int l,int r,int rt)
{
    if(col[rt]!=-1)
    {
        if(!hash[col[rt]]) cnt++;
        hash[col[rt]]=1;
        return;
    }
    if(l==r) return;
    int m=(l+r)/2;
    query(l,m,rt<<1);
    query(m+1,r,rt<<1|1);
}
int Bin(int num,int R)
{
    int l=0,r=R-1,mid;
    while(l<=r)
    {
        mid=(l+r)/2;
        if(X[mid]==num) return mid;
        if(X[mid]<num) l=mid+1;
        else r=mid-1;
    }
}
int main()
{
   // freopen("cin.txt","r",stdin);
    int t,n;
    scanf("%d",&t);
    while(t--)
    {
        int nn=0,m=1;
        scanf("%d",&n);
        for(int i=0;i<n;i++) scanf("%d%d",&li[i],&ri[i]),X[nn++]=li[i],X[nn++]=ri[i];
        sort(X,X+nn);
        for(int i=1;i<nn;i++)  if(X[i]!=X[i-1]) X[m++]=X[i];
        for(int i=m-1;i>0;i--)if(X[i]!=X[i-1]+1)X[m++]=X[i-1]+1;
        sort(X,X+m);
        memset(col,-1,sizeof(col));
        for(int i=0;i<n;i++)
        {
            int l=Bin(li[i],m);
            int r=Bin(ri[i],m);
            update(l,r,i,0,m,1);
        }
        cnt=0;
        memset(hash , false , sizeof(hash));
        query(0,m,1);
        printf("%d\n",cnt);
//        scanf("%d",&n);
//        int nn = 0;
//        for (int i = 0 ; i < n ; i ++) {
//            scanf("%d%d",&li[i] , &ri[i]);
//            X[nn++] = li[i];
//            X[nn++] = ri[i];
//        }
//        sort(X , X + nn);
//        int m = 1;
//        for (int i = 1 ; i < nn; i ++) {
//            if (X[i] != X[i-1]) X[m ++] = X[i];
//        }
//        for (int i = m - 1 ; i > 0 ; i --) {
//            if (X[i] != X[i-1] + 1) X[m ++] = X[i-1] + 1;
//        }
//        sort(X , X + m);
//        memset(col , -1 , sizeof(col));
//        for (int i = 0 ; i < n ; i ++) {
//            int l = Bin(li[i] , m);
//            int r = Bin(ri[i] , m);
//            update(l , r , i , 0 , m , 1);
//        }
//        cnt = 0;
//        memset(hash , false , sizeof(hash));
//        query(0 , m , 1);
//        printf("%d\n",cnt);
    }
    return 0;
}

纠结于更新操作的具体写法，发现如下写也是对的

/*******************
poj2528
2016.3.7
1088K	79MS	C++	2669B
*******************/
#include <cstdio>
#include <algorithm>
#include<cstring>
using namespace std;
#define lson l , m , rt << 1
#define rson m + 1 , r , rt << 1 | 1
const int maxn = 21111;
bool hash[maxn];
int li[maxn] , ri[maxn];
int X[maxn*3];
int col[maxn<<4];
int cnt;
void push_down(int rt)
{
    if(col[rt]!=-1)
    {
        col[rt<<1]=col[rt<<1|1]=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;
    }
    push_down(rt);
    int mid=(L+R)/2;
    if(r<=mid)update(l,r,c,L,mid,rt<<1);
    else if(l>mid) update(l,r,c,mid+1,R,rt<<1|1);///
    else
    {
        update(l,mid,c,L,mid,rt<<1);
        update(mid+1,r,c,mid+1,R,rt<<1|1);
    }
}
void query(int l,int r,int rt)
{
    if(col[rt]!=-1)
    {
        if(!hash[col[rt]]) cnt++;
        hash[col[rt]]=1;
        return;
    }
    if(l==r) return;
    int m=(l+r)/2;
    query(l,m,rt<<1);
    query(m+1,r,rt<<1|1);
}
int Bin(int num,int R)
{
    int l=0,r=R-1,mid;
    while(l<=r)
    {
        mid=(l+r)/2;
        if(X[mid]==num) return mid;
        if(X[mid]<num) l=mid+1;
        else r=mid-1;
    }
}
int main()
{
   // freopen("cin.txt","r",stdin);
    int t,n;
    scanf("%d",&t);
    while(t--)
    {
        int nn=0,m=1;
        scanf("%d",&n);
        for(int i=0;i<n;i++) scanf("%d%d",&li[i],&ri[i]),X[nn++]=li[i],X[nn++]=ri[i];
        sort(X,X+nn);
        for(int i=1;i<nn;i++)  if(X[i]!=X[i-1]) X[m++]=X[i];
        for(int i=m-1;i>0;i--)if(X[i]!=X[i-1]+1)X[m++]=X[i-1]+1;
        sort(X,X+m);
        memset(col,-1,sizeof(col));
        for(int i=0;i<n;i++)
        {
            int l=Bin(li[i],m);
            int r=Bin(ri[i],m);
            update(l,r,i,0,m,1);
        }
        cnt=0;
        memset(hash , false , sizeof(hash));
        query(0,m,1);
        printf("%d\n",cnt);
//        scanf("%d",&n);
//        int nn = 0;
//        for (int i = 0 ; i < n ; i ++) {
//            scanf("%d%d",&li[i] , &ri[i]);
//            X[nn++] = li[i];
//            X[nn++] = ri[i];
//        }
//        sort(X , X + nn);
//        int m = 1;
//        for (int i = 1 ; i < nn; i ++) {
//            if (X[i] != X[i-1]) X[m ++] = X[i];
//        }
//        for (int i = m - 1 ; i > 0 ; i --) {
//            if (X[i] != X[i-1] + 1) X[m ++] = X[i-1] + 1;
//        }
//        sort(X , X + m);
//        memset(col , -1 , sizeof(col));
//        for (int i = 0 ; i < n ; i ++) {
//            int l = Bin(li[i] , m);
//            int r = Bin(ri[i] , m);
//            update(l , r , i , 0 , m , 1);
//        }
//        cnt = 0;
//        memset(hash , false , sizeof(hash));
//        query(0 , m , 1);
//        printf("%d\n",cnt);
    }
    return 0;
}