題目鏈接:http://www.lightoj.com/volume_showproblem.php?problem=1044
題意:給你一個字符串,問最少分爲幾個迴文串?
思路:dp[i]表示從開頭到位置 i 的最優解,若[j,i]是迴文串,則dp[i] = min(dp[i],dp[j-1] +1)
代碼:
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <string.h>
#include <string>
using namespace std;
char s[1010];
int dp[1010];
bool is_ok(char a[], int L, int R)
{
while (L <= R)
{
if (a[L] != a[R])
{
return false;
}
L++;
R--;
}
return true;
}
int main()
{
int T;
scanf("%d", &T);
for (int t = 1;t <= T;t++)
{
memset(dp, 0, sizeof(dp));
scanf("%s", s + 1);
int len = strlen(s+1);
for (int i = 1;i <= len;i++)
{
dp[i] = 1010;
for (int j = 1;j <= i;j++)
{
if (is_ok(s, j, i))
{
if (j == 1)
dp[i] = 1;
else
dp[i] = min(dp[i], dp[j - 1] + 1);
}
}
}
printf("Case %d: %d\n", t, dp[len]);
}
return 0;
}