BestCoder Round #82 (div.1) 1002 HDU 5677 dp-类似多重背包的思想

链接：戳这里

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
 

题意：

给出n个串,从这n个串中选出K个回文子串,使得选出的K个回文子串的长度总和为L,满足条件输出True ,否则False

思路：

处理出长度为i的回文子串的个数(1<=i<=L)  num[i]

设置dp状态：dp[i][j][k]  表示当前执行了i次取舍操作之后,能否达到选取j个回文子串,长度为k,能达到状态dp[i][j][k]=1，否则为0

由于每次设置的dp状态只跟上一层的有关 ,可以使用滚动数组,这样就消去了一维i

dp[2][i][j] 表示当前选了i个回文子串使得长度为j的状态能否达到,而当前的状态只需要通过上一层转移即可

那么我们再来分析怎么实现dp状态的转移

1：dp[now][0][0]=1

2:for(1<=i<=L)  枚举长度为i的回文子串去计算贡献

3:   for(1<=l<=L)  表示当前长度为l

4:      for(1<=j<=K) 表示当前的回文子串个数为j

5：     for(1<=k<=num[i])  枚举每个长度为i的回文子串选取的个数

6: k+j<=K && l+k*i<=L  边界条件

dp[now][k+j][l+k*i] | =dp[last][k][j]  当前的回文子串个数为k+j,长度为l+k*i的状态需要从上一层的回文子       串个数为k,长度为j抑或得到

其实我现在都好不是很能理解,可能我需要磨一磨具体的多重背包的推导过程看看

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<vector>
#include <ctime>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<iomanip>
#include<cmath>
#define mst(ss,b) memset((ss),(b),sizeof(ss))
#define maxn 0x3f3f3f3f
#define MAX 1000100
///#pragma comment(linker, "/STACK:102400000,102400000")
typedef long long ll;
typedef unsigned long long ull;
#define INF (1ll<<60)-1
using namespace std;
int n,K,L;
int dp[2][110][110];
int num[110];
char s[110];
bool pd(int l,int r){
    while(l<=r){
        if(s[l]==s[r]){
            l++;
            r--;
        } else return false;
    }
    return true;
}
int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d%d",&n,&K,&L);
        mst(num,0);
        for(int i=1;i<=n;i++){
            scanf("%s",&s);
            for(int j=0;j<strlen(s);j++){
                for(int k=j;k<strlen(s);k++){
                    if(pd(j,k)) num[k-j+1]++;
                }
            }
        }
        int now=0,last=1;
        dp[now][0][0]=1;
        for(int i=1;i<=L;i++){
            swap(now,last);
            mst(dp[now],0);
            for(int l=0;l<=L;l++){
                for(int j=0;j<=K;j++){
                    for(int k=0;k<=num[i];k++){
                        if(k+j>K || l+k*i>L) continue;
                        dp[now][k+j][l+k*i]|=dp[last][j][l];
                    }
                }
            }
        }
        if(dp[now][K][L]) printf("True\n");
        else printf("False\n");
    }
    return 0;
}