uva_10617_Again Palindrome(普通DP, 记忆化搜索)

这里的阶段就是按照字符串的顺序的方向进行规划
每个阶段的状态dp[i][j] 表示字符串中从第i个字符到第j个字符回文串的个数
dp[i+1][j-1]+cost   str[i] = str[j]
状态转移: dp[i][j] =
dp[i+1][j-1]
dp[i][j] 是枚举i到j的所有可能，还有的是要注意重复加

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

#define MAXN    61

char str[MAXN];
int  added[MAXN][MAXN];
long long dp[MAXN][MAXN];

long long dfs(const int &l, const int &r)
{
if( l >= r ) {
return dp[l][r] = 0;
}
if( -1 != dp[l][r] ) {
return dp[l][r];
}
long long rst(0);
for(int i = l; i <= r; i ++) {
for(int j = i+1; j <= r; j ++) {
dp[i+1][j-1] = (-1 == dp[i+1][j-1])? dfs(i+1, j-1) : dp[i+1][j-1];
if( str[i] == str[j] ) {
rst += dp[i+1][j-1]+(j-i-1)+1;
}
else if( !added[i+1][j-1] ) {
rst += dp[i+1][j-1];
}
}
}
added[l][r] = 1;
return dp[l][r] = rst;
}

int main(int argc, char const *argv[])
{
#ifndef ONLINE_JUDGE
freopen("test.in", "r", stdin);
#endif
int cas;
scanf("%d", &cas);
for( ; cas; cas --) {
scanf("%s", str);
memset(dp, -1, sizeof(dp)); memset(added, 0, sizeof(added));
printf("%lld\n", dfs(0, strlen(str)-1)+strlen(str));
}
return 0;
}