codeforces 666A （DP）

题目链接：http://codeforces.com/problemset/problem/666/A

思路：dp[i][0]表示第a[i-1]~a[i]组成的字符串是否可行，dp[i][1]表示第a[i-2]~a[i]组成的字符串是否可行，显然dp[len-2][0(1)]必定不可行。

转移方程：

dp[i][0] = dp[i+3][1] || dp[i+2][0] && (tmp1 != tmp2);

dp[i][1] = dp[i+2][0] || dp[i+3][1] && (tmp1 != tmp2);

#include<bits/stdc++.h>
using namespace std;
const int N = 1e4 + 10;
typedef long long ll;
char a[N];
string ans[N<<1];
string tmp1 ,tmp2;
bool dp[N][2];
int cur ,num = 1;
int main()
{
    scanf("%s",a);
    int len = strlen(a);
    if(len <= 6)
    {
        printf("0
");
        return 0;
    }
    dp[len-1][0] = 1;
    tmp1 = "";
    tmp1 += a[len-2];
    tmp1 += a[len-1];
    ans[cur++] = tmp1;
    if(len > 7)
    {
        dp[len-1][1] = 1;
        tmp1 = "";
        tmp1 += a[len-3];
        tmp1 += a[len-2];
        tmp1 += a[len-1];
        ans[cur++] = tmp1;
    }
    for(int i = len - 3 ;i >= 6 ;i--)
    {
        tmp1 = "";
        tmp1 += a[i-1];
        tmp1 += a[i];
        tmp2 = "";
        tmp2 += a[i+1];
        tmp2 += a[i+2];
        dp[i][0] = dp[i+3][1] || dp[i+2][0] && (tmp1 != tmp2);
        if(dp[i][0])
            ans[cur++] = tmp1;

        if(i == 6)
            continue;

        tmp1 = "";
        tmp1 += a[i-2];
        tmp1 += a[i-1];
        tmp1 += a[i];
        tmp2 = "";
        tmp2 += a[i+1];
        tmp2 += a[i+2];
        tmp2 += a[i+3];
        dp[i][1] = dp[i+2][0] || dp[i+3][1] && (tmp1 != tmp2);
        if(dp[i][1])
            ans[cur++] = tmp1;
    }
    sort(ans ,ans + cur);//排序
    for(int i = 1 ; i < cur ; i++)
    {
        if(ans[i] != ans[i-1])//并去重
            num++;
    }
    printf("%d
",num);
    cout<<ans[0]<<"
";
    for(int i = 1 ; i < cur ; i++)
    {
        if(ans[i] != ans[i-1])
            cout<<ans[i]<<"
";
    }
    return 0;
}