【 2013 Multi-University Training Contest 4 】

HDU 4632 Palindrome subsequence

dp[x][y]表示区间[x,y]构成回文串的方案数。

若str[x]==str[y]，dp[x][y]=dp[x+1][y]+dp[x][y-1]-dp[x+1][y-1]+(dp[x+1][y-1]+1)=dp[x+1][y]+dp[x][y-1]+1。

若str[x]!=str[y]，dp[x][y]=dp[x+1][y]+dp[x][y-1]-dp[x+1][y-1]。

 1 #include<cstdio>
 2 #include<cstring>
 3 #define MAXN 1010
 4 #define MOD 10007
 5 char str[MAXN];
 6 int dp[MAXN][MAXN];
 7 int dfs(int x, int y) {
 8     if (dp[x][y] == -1) {
 9         if (str[x] != str[y]) {
10             dp[x][y] = dfs(x + 1, y) + dfs(x, y - 1) - dfs(x + 1, y - 1);
11         } else {
12             dp[x][y] = dfs(x + 1, y) + dfs(x, y - 1) + 1;
13         }
14     }
15     dp[x][y] %= MOD;
16     return dp[x][y];
17 }
18 int main() {
19     int T;
20     int ca = 1;
21     int len;
22     int i;
23     scanf("%d", &T);
24     while (T--) {
25         memset(dp, -1, sizeof(dp));
26         scanf(" %s", str);
27         len = strlen(str);
28         for (i = 0; i < len; i++) {
29             dp[i][i] = 1;
30         }
31         for (i = 1; i < len; i++) {
32             if (str[i] == str[i - 1]) {
33                 dp[i - 1][i] = 3;
34             } else {
35                 dp[i - 1][i] = 2;
36             }
37         }
38         printf("Case %d: %d\n", ca++, (dfs(0, len - 1) + MOD) % MOD);
39     }
40     return 0;
41 }

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