2013 Multi-University Training Contest 4

HDU-4632 Palindrome subsequence

题意：给定一个字符串，长度最长为1000，问该串有多少个回文子串。

分析：设dp[i][j]表示从 i 到 j 有多少个回文子串，则有动态规划方程：

str[i] != str[j]：dp[i][j] = dp[i+1][j] + dp[i][j-1] - dp[i+1][j-1];
str[i]  = str[j]：dp[i][j] = dp[i+1][j] + dp[i][j-1] + 1.

#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const int mod = 10007;
const int N = 1005;
char str[N];
int f[N][N];

int solve(int len) {
    memset(f, 0, sizeof (f));
    for (int i = 0; i < len; ++i) {
        f[i][i] = 1;
    }
    for (int k = 2; k <= len; ++k) { // 枚举长度 
        for (int i = 0, j; i < len && (j=i+k-1) < len; ++i) {
            if (str[i] == str[j]) {
                f[i][j] = (f[i][j-1] + f[i+1][j] + 1) % mod;
            } else {
                f[i][j] = ((f[i][j-1] + f[i+1][j] - f[i+1][j-1]) % mod + mod) % mod;
            }
        }
    }
    return f[0][len-1];
}

int main() {
    int T, ca = 0;
    scanf("%d", &T);
    while (T--) {
        scanf("%s", str);
        int len = strlen(str);
        printf("Case %d: %d\n", ++ca, solve(len));
    }
    return 0;    
}

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