Intervals+线段树+贪心+poj

Intervals

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 21403		Accepted: 8053

Description

You are given n closed, integer intervals [ai, bi] and n integers c1, ..., cn. 
Write a program that: 
reads the number of intervals, their end points and integers c1, ..., cn from the standard input, 
computes the minimal size of a set Z of integers which has at least ci common elements with interval [ai, bi], for each i=1,2,...,n, 
writes the answer to the standard output. 

Input

The first line of the input contains an integer n (1 <= n <= 50000) -- the number of intervals. The following n lines describe the intervals. The (i+1)-th line of the input contains three integers ai, bi and ci separated by single spaces and such that 0 <= ai <= bi <= 50000 and 1 <= ci <= bi - ai+1.

Output

The output contains exactly one integer equal to the minimal size of set Z sharing at least ci elements with interval [ai, bi], for each i=1,2,...,n.

Sample Input

5
3 7 3
8 10 3
6 8 1
1 3 1
10 11 1

Sample Output

6

解决方案：这题可用差分约束法，也可用线段树+贪心的方法。第二种是这样子处理的，先将每个区间按有段坐标从小到大排序，然后每从左到右选取区间进行选点，选点要从该区间右端开始选。每次先看区间已选的点有多少，然后从右端开始把不够的补上。这个过程可用线段树维护，把效率提高了。通过这道题，我对线段树的使用有了进一步的理解。

code:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define MMAX 50005
using namespace std;
int sum[MMAX*4],setv[MMAX*4];
int t,_sum;
struct node
{
    int l,r,v;
} line[MMAX];
bool cmp(node a,node b)
{
    return a.r<b.r;
}
void build(int rt,int L,int R)
{
    if(L==R)
    {
        sum[rt]=1;
    }
    else
    {
        int M=(L+R)/2;
        build(rt*2,L,M);
        build(rt*2+1,M+1,R);
        sum[rt]=sum[rt*2]+sum[rt*2+1];
    }
}
void update(int rt,int L,int R,int ql,int qr)
{
    if(t==0) return ;
 //   if(sum[rt]==0) return ;
    if(ql<=L&&qr>=R&&sum[rt]&&t>=sum[rt])
    {
        t-=sum[rt];
        sum[rt]=0;
        return ;
    }
    if(L==R) return ;
    if(sum[rt]==0)
    {
        int lc=rt*2,lr=rt*2+1;
        sum[lr]=sum[lc]=0;
    }
    int M=(L+R)/2;
    if(qr>M) update(rt*2+1,M+1,R,ql,qr);
    if(ql<=M) update(rt*2,L,M,ql,qr);
    sum[rt]=sum[rt*2+1]+sum[rt*2];

}
void query(int rt,int L,int R,int ql,int qr)
{
    if(sum[rt]==0) return ;
    if(ql<=L&&qr>=R)
    {
        _sum+=sum[rt];
    }
    else
    {
         if(sum[rt]==0)
        {
        int lc=rt*2,lr=rt*2+1;
        sum[lr]=sum[lc]=0;
        }
        int M=(L+R)/2;
        if(ql<=M) query(rt*2,L,M,ql,qr);
        if(qr>M) query(rt*2+1,M+1,R,ql,qr);
    }

}
int main()
{
    int n;
    while(~scanf("%d",&n))
    {

        memset(setv,-1,sizeof(setv));
        int Max=0;
        for(int i=0; i<n; i++)
        {
            scanf("%d%d%d",&line[i].l,&line[i].r,&line[i].v);
            Max=max(Max,line[i].r);
        }
        build(1,0,Max);
        sort(line,line+n,cmp);
        for(int i=0; i<n; i++)
        {
            _sum=0;
            query(1,0,Max,line[i].l,line[i].r);
           // cout<<_sum<<endl;
            _sum=line[i].v-(line[i].r-line[i].l+1-_sum);

            if(_sum<=0) continue;
            t=_sum;
            update(1,0,Max,line[i].l,line[i].r);

        }
        _sum=0;
        query(1,0,Max,0,Max);
        printf("%d\n",Max-_sum+1);
    }
    return 0;
}