HDOJ 5677 ztr loves substring(Manacher+背包型DP)

来源：互联网发布：寻侠九宫突破数据编辑：程序博客网时间：2024/06/06 04:40

题意

问能不能从给出的字符串中找到k个回文子串能组成总长度为l的串。

思路

先Manacher预处理出所有长度的回文串的个数，然后就变成一个多重背包了，即dp[i][j][l]表示枚举到长度为i的回文串时已经取了j个串总长度为l的情况是否存在，因为长度为i的回文串有很多，一个一个枚举肯定就T了，我们就用多重背包那种做法把数量变成二进制来处理就行了，总复杂度L∗K∗K)∗log(∑cnt[i])。
马拉车的时候str数组忘了开两倍wa了n次。。。mdzz

代码

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <stack>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>#include <time.h>using namespace std;#define LL long long#define Lowbit(x) ((x)&(-x))#define lson l, mid, rt << 1#define rson mid + 1, r, rt << 1|1#define MP(a, b) make_pair(a, b)const int INF = 0x3f3f3f3f;const int MOD = 1000000007;const int maxn = 1e5 + 10;const double eps = 1e-8;const double PI = acos(-1.0);typedef pair<int, int> pii;char s[210], str[210];int cnt[210], p[210];int dp[2][110][110];void Manacher(int len){    int mx = 0, id = 0;    for (int i = 0; i < len; i++)    {        if (mx > i) p[i] = min(p[2*id-i], mx - i);        else p[i] = 1;        while (str[i-p[i]] == str[i+p[i]]) p[i]++;        if (p[i] + i > mx)            mx = i + p[i], id = i;    }}int init(){    int len = strlen(s);    int l = 0;    str[l++] = '$';    str[l++] = '#';    for (int i = 0; i < len; i++)    {        str[l++] = s[i];        str[l++] = '#';    }    str[l] = 0;    return l;}int main(){    //freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);    int T;    cin >> T;    while (T--)    {        int n, K, L;        scanf("%d%d%d", &n, &K, &L);        memset(cnt, 0, sizeof(cnt));        for (int i = 0; i < n; i++)        {            scanf("%s", s);            int len = init();            Manacher(len);            //for (int j = 0; j < len; j++) printf("%d %d %c %d\n", i, j, str[j], p[j]);            for (int j = 0; j < len; j++)                if (p[j] > 1) cnt[p[j]-1]++;        }        for (int i = L; i > 0; i--)            cnt[i] += cnt[i+2];        memset(dp, 0, sizeof(dp));        int now = 0, la = 1;        dp[0][0][0] = 1;        for (int i = 1; i <= L; i++)        {            now ^= 1;            memset(dp[now], 0, sizeof(dp[now]));            for (int j = 0; j <= K; j++)                for (int o = 0; o <= L; o++)                    for (int num = 0; num <= cnt[i] && num + j <= K && num * i + o <= L; num++)                        dp[now][num+j][num*i+o] |= dp[now^1][j][o];        }        puts(dp[now][K][L] ? "True" : "False");    }    return 0;}

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