[HDU5677]ztr loves substring

ztr loves substring

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Problem Description

ztr love reserach substring.Today ,he has n string.Now ztr want to konw,can he take out exactly k palindrome from all substring of these n string,and thrn sum of length of these k substring is L.for example string "yjqqaq".this string contains plalindromes:"y","j","q","a","q","qq","qaq".so we can choose "qq" and "qaq".

Input

The first line of input contains an positive integer T(T<=10) indicating the number of test cases.
For each test case:
First line contains these positive integer N(1<=N<=100),K(1<=K<=100),L(L<=100).
The next N line,each line contains a string only contains lowercase.Guarantee even length of string won't more than L.

Output

For each test,Output a line.If can output "True",else output "False".

Sample Input

3

2 3 7

yjqqaq

claris

2 2 7

popoqqq

fwwf

1 3 3

aaa

Sample Output

False

True

True

 

题解：

（多重背包都快不会打了我已经完蛋了）

这道题相对来说还是比较裸的。先用manacher计算出每种长度回文串出现的个数。

在将长度视为物品跑一个多重背包即可。

关于这里的背包，我们可以设f[i][j]为一个bool滚动数组，表示构造长度为i的串用了j个子串能不能成立。

那么显然3层循环即可。我打的是二进制分解然后01背包（其实是因为单调队列不会）

代码见下：

#include<cstdio>
#include<cstring>
using namespace std;
const int N=110;
int n,t,m,k,ct,mx,l;
int r[N<<1],f[N][N],vis[N];
char s[N<<1],str[N];
inline int max(int a,int b){return a>b?a:b;}
inline int min(int a,int b){return a<b?a:b;}
inline void manacher()
{
    memset(s,0,sizeof(s));
    n=0;m=strlen(str);
    s[n++]=1;s[n++]=2;
    for(int i=0;i<m;i++)s[n++]=str[i],s[n++]=2;
    memset(r,0,sizeof(r));ct=mx=0;
    for(int i=0;i<n;i++)
    {
        if(i<mx)r[i]=min(mx-i,r[2*ct-i]);
        else r[i]=1;
        while(0<=i-r[i]&&i+r[i]<n&&s[i-r[i]]==s[i+r[i]])r[i]++;
        if(i+r[i]>mx)ct=i,mx=i+r[i];
        int j=r[i]-1;
        while(j>=0)vis[j]++,j-=2;
    }
}
int c[N*N],v[N*N],cnt;
inline bool backpack()
{
    memset(f,0,sizeof(f));
    for(int i=1;i<=l;i++)
    {
        int tmp=1;
        while(vis[i]>=tmp)
            c[++cnt]=tmp*i,v[cnt]=tmp,vis[i]-=tmp,tmp<<=1;
        if(vis[i])
            c[++cnt]=vis[i]*i,v[cnt]=vis[i];
    }
    f[0][0]=1;
    for(int u=1;u<=cnt;u++)
        for(int i=l;i>=c[u];i--)
            for(int j=k;j>=v[u];j--)
                f[i][j]|=f[i-c[u]][j-v[u]];
    return f[l][k];
}
int main()
{
    int cnt;scanf("%d",&cnt);
    while(cnt--)
    {
        scanf("%d%d%d",&t,&k,&l);
        memset(vis,0,sizeof(vis));
        for(int i=1;i<=t;i++)
            scanf("%s",str),manacher();
        if(backpack())printf("True
");
        else printf("False
");
    }
}