uva_10617_Again Palindrome(普通DP, 记忆化搜索)

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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    61char 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;}