Palindrome subsequence HDU - 4632 (区间dp)
题目链接
题目大意:求字符串有多少回文子序列。(这里的子序列可以不连续)
input:
4
a
aaaaa
goodafternooneveryone
welcometoooxxourproblems
output:
Case 1: 1
Case 2: 31
Case 3: 421
Case 4: 960
#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;
const int maxn = 1e3 + 100;
const int inf = 0x3f3f3f3f;
const int mod = 10007;
char str[maxn];
int dp[maxn][maxn];
int main()
{
int t, cas = 1;
scanf("%d", &t);
while(t--)
{
scanf("%s", str);
int len = strlen(str);
memset(dp, 0, sizeof(dp));
for(int i = 0; i < len; i++) {
dp[i][i] = 1;
}
for(int i = 1; i < len; i++) {
for(int j = i - 1; j >= 0; j--) {
dp[j][i] = (dp[j + 1][i] + dp[j][i - 1] - dp[j + 1][i - 1] + mod) % mod; //小的容斥
if(str[i] == str[j]) dp[j][i] = (dp[j][i] + dp[j + 1][i - 1] + 1) % mod;
}
}
printf("Case %d: %d\n", cas++, dp[0][len - 1]);
}
}