hdu 4352 XHXJ's LIS

http://acm.hdu.edu.cn/showproblem.php?pid=4352

 

题意:

设一个数的LIS为该数各位拆开来后的最长上升子序列。例如1324的LIS为3。
求l~r中LIS为k的数的个数。
 
如果能想办法将“最长上升子序列”这个状态压缩为s,那就可以数位dp
dp[i][s][k]表示前i位,状态为s,LIS长度为k的方案数
 
如何压缩状态?
考虑nlogn求最长上升子序列方法:
f[i] 表示长为i的上升序列,最后一个数的最小值
f数组是递增的,且本题中只可能是0~9
令状态s表示 0~9是否出现在了f数组中
 
转移的时候,枚举下一位数x
从s中删掉最小的大于等于x的数,把x加进去,就得到了新的状态s
 
#include<cstdio>
#include<cstring>
#include<iostream>

using namespace std;

typedef long long LL;

int k;
int a[20];

LL dp[20][(1<<10)+2][11];

template<typename T>
void read(T &x)
{
    x=0; char c=getchar();
    while(!isdigit(c)) c=getchar();
    while(isdigit(c)) { x=x*10+c-'0'; c=getchar(); }
}

int trans(int s,int num,int & mx)
{
    int d=-1;
    for(int i=num;i<=9;++i)
        if(s&(1<<i)) { d=i; break; }
    if(d==-1)
    {
        mx++;
        s|=1<<num;
    }
    else
    {
        s^=1<<d;
        s|=1<<num;
    }
    return s;
}

LL dfs(int dep,int s,int len,bool lim,bool zero)
{
    if(!lim && dp[dep][s][k]!=-1) return dp[dep][s][k];
    if(!dep) return len==k;
    LL tot=0;
    int up=lim ? a[dep] : 9;
    int news,newlen;
    for(int i=0;i<=up;++i)
    {
        newlen=len;
        if(!i && zero) news=0;
        else news=trans(s,i,newlen);
        tot+=dfs(dep-1,news,newlen,lim && i==up,zero && !i);
    }
    if(!lim) dp[dep][s][k]=tot;
    return tot;
}

LL solve(LL m)
{
    int n=0;
    while(m) a[++n]=m%10,m/=10;
//    memset(dp,-1,sizeof(dp));
    return dfs(n,0,0,true,true);
}

int main()
{
    int T;
    LL L,R;
    read(T);
    memset(dp,-1,sizeof(dp));
    for(int t=1;t<=T;++t)
    {
        read(L); read(R); read(k);
        printf("Case #%d: ",t);
        cout<<solve(R)-solve(L-1)<<'\n';
    }
}

 

XHXJ's LIS

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3638    Accepted Submission(s): 1519


Problem Description
#define xhxj (Xin Hang senior sister(学姐))
If you do not know xhxj, then carefully reading the entire description is very important.
As the strongest fighting force in UESTC, xhxj grew up in Jintang, a border town of Chengdu.
Like many god cattles, xhxj has a legendary life:
2010.04, had not yet begun to learn the algorithm, xhxj won the second prize in the university contest. And in this fall, xhxj got one gold medal and one silver medal of regional contest. In the next year's summer, xhxj was invited to Beijing to attend the astar onsite. A few months later, xhxj got two gold medals and was also qualified for world's final. However, xhxj was defeated by zhymaoiing in the competition that determined who would go to the world's final(there is only one team for every university to send to the world's final) .Now, xhxj is much more stronger than ever,and she will go to the dreaming country to compete in TCO final.
As you see, xhxj always keeps a short hair(reasons unknown), so she looks like a boy( I will not tell you she is actually a lovely girl), wearing yellow T-shirt. When she is not talking, her round face feels very lovely, attracting others to touch her face gently。Unlike God Luo's, another UESTC god cattle who has cool and noble charm, xhxj is quite approachable, lively, clever. On the other hand,xhxj is very sensitive to the beautiful properties, "this problem has a very good properties",she always said that after ACing a very hard problem. She often helps in finding solutions, even though she is not good at the problems of that type.
Xhxj loves many games such as,Dota, ocg, mahjong, Starcraft 2, Diablo 3.etc,if you can beat her in any game above, you will get her admire and become a god cattle. She is very concerned with her younger schoolfellows, if she saw someone on a DOTA platform, she would say: "Why do not you go to improve your programming skill". When she receives sincere compliments from others, she would say modestly: "Please don’t flatter at me.(Please don't black)."As she will graduate after no more than one year, xhxj also wants to fall in love. However, the man in her dreams has not yet appeared, so she now prefers girls.
Another hobby of xhxj is yy(speculation) some magical problems to discover the special properties. For example, when she see a number, she would think whether the digits of a number are strictly increasing. If you consider the number as a string and can get a longest strictly increasing subsequence the length of which is equal to k, the power of this number is k.. It is very simple to determine a single number’s power, but is it also easy to solve this problem with the numbers within an interval? xhxj has a little tired,she want a god cattle to help her solve this problem,the problem is: Determine how many numbers have the power value k in [L,R] in O(1)time.
For the first one to solve this problem,xhxj will upgrade 20 favorability rate。
 

 

Input
First a integer T(T<=10000),then T lines follow, every line has three positive integer L,R,K.(
0<L<=R<263-1 and 1<=K<=10).
 

 

Output
For each query, print "Case #t: ans" in a line, in which t is the number of the test case starting from 1 and ans is the answer.
 

 

Sample Input
1 123 321 2
 

 

Sample Output
Case #1: 139
 

 

Author
peterae86@UESTC03
 
 
 
posted @ 2018-02-20 16:14  TRTTG  阅读(265)  评论(0编辑  收藏  举报