hdu 6003 Problem Buyer（贪心）

Problem Buyer

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 195    Accepted Submission(s): 60

Problem Description

TopSetter is an organization that creates problems. They’ve prepared N problems with estimated difficulty score in range [ Ai,Bi ]. TopHoster would like to host a contest consisting of M problems.
The ith problem should be of difficulty score Ci. The ith problem from TopSetter can be used in the contest if and only if its estimated difficulty score range [Ai,Bi] covers the difficulty score c of its target problem in the contest, i.e. Ai≤c≤Bi . Hosting a contest with M problems needs tohave M distinct problems which satisfy the required difficulty scores for each problem.
Unfortunately, TopSetter doesn’t provide a service to buy specific problems. You can only request a problem set containing K problems and they will give you K distinct problems from all the N problems, but you don’t know which problems will be given.
As TopSetter is the only problem provider for TopHoster, TopHoster would like to know the least number K of problems they need to buy to make sure they can host a contest.

 

Input

The first line of the input gives the number of test cases, T. T test cases follow. Each test case starts with 2 integers, N and M. Then N lines follow, each line consists of 2 integers representing the difficulty score range of the ith problem, AiandBi . The last line of each test case consists of M integers representing the target difficulty scores of the M problems Ci .

 

Output

For each test case, output one line containing “Case #x: y”, where x is the test case number (starting from 1) and y is the least number of problems which the TopHoster needs to buy.
Output “IMPOSSIBLE!” if it’s impossible.

limits

∙1≤T≤100.
∙1≤N,M≤105.
∙1≤Ai≤Bi≤109.
∙1≤Ci≤109.

 

Sample Input

33 11 42 35 633 21 103 47 94 83 31 25 68 91 5 10

 

Sample Output

Case #1: 2Case #2: 2Case #3: IMPOSSIBLE!

读了好久连题目意思都读不懂。。。

真是一道超级难想的贪心题

题意：给出n个区间，m个数，让你任意选k个数，使一定能包含这m个数，一个区间只能包含一个数 求最小的k

解：难就难在 任意 二字 解题思路是 求出一个数的所有满足区间 那么这个区间取一个数即可包含这个数 那么ans=n-size+1

解题报告：http://www.cnblogs.com/xiaochaoqun/p/7243606.html

#include <iostream>#include <stdio.h>#include <string.h>#include <stack>#include <queue>#include <map>#include <set>#include <vector>#include <math.h>#include <bitset>#include <algorithm>#include <climits>#include <bits/stdc++.h>using namespace std;const int N =1e5+10;typedef long long LL;const LL mod = 1000000007;typedef pair<int,int>pi;struct node{    int l, r;    bool operator <(const node &A)const    {        if(l!=A.l) return l<A.l;        else return r>A.r;    }}p[N];int c[N];priority_queue<int,vector<int>,greater<int> >q;int main(){    int t, ncase=1;    scanf("%d", &t);    while(t--)    {        int n, m;        scanf("%d %d", &n, &m);        for(int i=0;i<n;i++) scanf("%d %d", &p[i].l,&p[i].r);        for(int i=0;i<m;i++) scanf("%d", &c[i]);        sort(p,p+n);        sort(c,c+m);        int ans=-1;        while(!q.empty()) q.pop();        int pos=0;        for(int i=0;i<m;i++)        {            while(pos<n&&p[pos].l<=c[i])            {                if(p[pos].r>=c[i])                {                    q.push(p[pos].r);                }                pos++;            }            while(!q.empty()&&q.top()<c[i]) q.pop();            if(q.empty())            {                ans=-1;                break;            }            ans=max(ans,n-(int)q.size()+1);            q.pop();        }        if(ans==-1) printf("Case #%d: IMPOSSIBLE!\n",ncase++);        else printf("Case #%d: %d\n",ncase++,ans);    }    return 0;}