hdu 5677 ztr loves substring 多重背包

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 178    Accepted Submission(s): 93

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

 

Source

BestCoder Round #82 (div.2)

思路:二维费用的

#include <bits/stdc++.h>
using namespace std;
const int N = ;
const int INF = 0x3f3f3f3f;
int Ma[N << ], Mp[N << ];
char s[N];
int cnt[N];

void manacher(char s[], int len) {
    int l = ;
    Ma[l++] = '$';
    Ma[l++] = '#';
    for(int i = ; i < len; ++i)
    {
        Ma[l++] = s[i];
        Ma[l++] = '#';
    }
    Ma[l] = ;
    int mx = , id = ;
    for(int i = ; i < l; ++i)
    {
        Mp[i] = mx > i ? min(Mp[ * id - i], mx - i) : ;
        while(Ma[ i + Mp[i] ] == Ma[ i - Mp[i] ]) Mp[i]++;
        if(i + Mp[i] > mx) mx = i + Mp[i], id = i;
    }
    for(int i = ; i < l; ++i) if(Mp[i] > ) {
        int p;
        if(Mp[i] & ) p = ; else p = ;
        for(int j = p; j < Mp[i]; j += ) cnt[j]++;
    }
}
int dp[N][N];
void Dp1(int v, int k, int l) {
    for(int i = v; i <= l; ++i)
    for(int j = ; j <= k; ++j)
    if(dp[i - v][j - ]) dp[i][j] = ;
}
void Dp2(int v, int num, int k, int l) {
    for(int i = l; i >= v; --i)
    for(int j = k; j >= num; --j)
    if(dp[i - v][j - num]) dp[i][j] = ;
}

void to(int x, int k, int l) {
    if(cnt[x] >= k && x * cnt[x] >= l) Dp1(x, k, l);
    else {
        int s = , tot = cnt[x];
        while(s < tot) {
            Dp2(x * s, s, k, l);
            tot -= s;
            s <<= ;
        }
        Dp2(tot * x, tot, k, l);
    }
}
bool solve(int k, int l) {
    memset(dp, , sizeof dp);
    dp[][] = ;
    for(int i = ; i <= ; ++i) if(cnt[i]) to(i, k, l);
    if(dp[l][k]) return true;
    return false;
}
int main() {
   // freopen("in", "r", stdin);
    int _; scanf("%d", &_);
    while(_ --) {
        int n, k, l;
        scanf("%d%d%d", &n, &k, &l);
        memset(cnt, , sizeof cnt);
        for(int i = ; i < n; ++i) {
            scanf("%s", s);
            int len = strlen(s);
            manacher(s, len);
        }
        if(solve(k, l)) puts("True");
        else puts("False");
    }
    return ;
}

,且二维都要求恰好装满,那么初始化[0][0]能满足，其它状态都不满足