Codeforces 557E Ann and Half-Palindrome DP+字典树

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求字典序第K大的半回文串，半回文串满足第一位和最后一位相等，第三位和倒数第三位相等。。。。先用dp求出所有的半回文串，然后插入到字典树中统计个数，根据个数求第K大。

#include #include char s[5010];int k;int dp[5010][5010];int ch[5010*5010/2][2];int sum[5010*5010/2];int val[5010*5010/2];int sz;void init() {    sz = 1;}void dfs(int u, int x, int n, int pos) {    if(n == x) {        return;    }    int c = s[x]-'a';    if(ch[u][c] == 0)    {        ch[u][c] = sz++;    }    int v = ch[u][c];    dfs(v, x+1, n, pos);    if(dp[pos][x] == 1) {        val[v] += 1;    }    sum[v] = sum[ch[v][0]] + sum[ch[v][1]] + val[v];}void insert(int pos) {    int u = 0, n = strlen(s);    dfs(u, pos, n, pos);}void find(int k) {    char ans[5010];    int t = 0;    int u = 0;    while(k > 0) {        int c = ch[u][0];        //System.out.println(sum[u] + " " + sum[ch[u][0]] + " " + sum[ch[u][1]] + " " + k);        int num = sum[u]-(sum[ch[u][0]]+sum[ch[u][1]]);        if(u == 0)            num = 0;        if(k <= num)            break;        k -= num;        if(sum[c] >= k) {            u = c;            ans[t++] = 'a';        }        else {            k -= sum[c];            u = ch[u][1];            ans[t++] = 'b';        }    }    ans[t] = 0;    printf("%s", ans);}void dfs2(int u) {    if(u == 0)        return;    dfs2(ch[u][0]);    dfs2(ch[u][1]);    sum[u] += sum[ch[u][0]] + sum[ch[u][1]];}int main() {    scanf("%s %d", s, &k);    int len = strlen(s);    for(int i = 0; i < len; i++) {        dp[i][i] = 1;    }    for(int i = 0 ; i+1 < len; i++) {        if(s[i] == s[i+1])            dp[i][i+1] = 1;    }    for(int i = 0 ; i+2 < len; i++) {        if(s[i] == s[i+2])            dp[i][i+2] = 1;    }    for(int i = 0 ; i+3 < len; i++) {        if(s[i] == s[i+3])            dp[i][i+3] = 1;    }    for(int i = 5; i <= len; i++) {        for(int j = 0; j+i <= len; j++) {            if(s[j] == s[j+i-1] && dp[j+2][j+i-3] == 1) {                dp[j][j+i-1] = 1;            }        }    }    init();    for(int i = 0; i < len; i++) {        insert(i);    }    //dfs2(ch[0][0]);    //dfs2(ch[0][1]);    find(k);}

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