HDU 6249
Alice’s Stamps
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 125 Accepted Submission(s): 34
There are N different kinds of stamps that exist in the world; they are numbered 1 through N. However, stamps are not sold individually; they must be purchased in sets. There are M different stamp sets available; the ith set contains the stamps numbered Li through Ri. The same stamp might appear in more than one set, and it is possible that one or more stamps are not available in any of the sets.
All of the sets cost the same amount; because Alice has a limited budget, she can buy at most K different sets. What is the maximum number of different kinds of stamps that Alice can get?
Each test case begins with a line containing three integers: N, M, and K: the number of different kinds of stamps available, the number of stamp sets available, and the maximum number of stamp sets that Alice can buy.
M lines follow; the ithoftheselinesrepresentsthei^{th} stamp set and contains two integers, Li and Ri, which represent the inclusive range of the numbers of the stamps available in that set.
1≤T≤100
1≤K≤M
1≤N,M≤2000
1≤Li≤Ri≤N
5 3 2
3 4
1 1
1 3
100 2 1
1 50
90 100
Case #2: 50
In sample case #1, Alice could buy the first and the third stamp sets, which contain the first four kinds
of stamp. Note that she gets two copies of stamp 3, but only the number of different kinds of stamps
matters, not the number of stamps of each kind.
In sample case #2, Alice could buy the first stamp set, which contains 50 different kinds of stamps.
题意:
有n种卡片1~n,m个卡包,每个卡包中包含一段连续的不重复的卡片,wen最多买k个卡包能够得到的最多的种类的卡片。
代码:
//n和k只有2000,可以用nk的方法,f[i][j]表示到达i位置使用了j个卡包的最大卡片种类数,先处理出来如果买了i卡片能够买到的左边的最远的那
//个卡片lef[i],这样就可以转移了,到达i位置考虑买还是不买,不买就由前一个转移,买就由lef[i]-1转移来。
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int INF=0x3f3f3f3f;
int t,n,m,k,f[][],lef[];
int main()
{
scanf("%d",&t);
for(int cas=;cas<=t;cas++){
memset(f,,sizeof(f));
memset(lef,INF,sizeof(lef));
scanf("%d%d%d",&n,&m,&k);
for(int i=;i<=m;i++){
int x,y;
scanf("%d%d",&x,&y);
for(int j=x;j<=y;j++) lef[j]=min(lef[j],x);
}
int ans=;
for(int i=;i<=n;i++){
for(int j=;j<=k;j++){
if(lef[i]==INF){
f[i][j]=f[i-][j];
continue;
}
int tmp=lef[i]-;
f[i][j]=max(f[i-][j],f[tmp][j-]+(i-tmp));
ans=max(ans,f[i][j]);
}
}
printf("Case #%d: %d\n",cas,ans);
}
return ;
}
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