UVA11584 划分成回文串

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=105116#problem/B

紫书275

题意:输入一个字符，最少能划分几个回文串

分析：预处理一下，判断i，j是否为回文串；动态分析求解，dp[i] = dp[i - 1] + 1,假设i单独成为一个回文串，然后在往前找，如果j到i是回文，dp[i] = min(dp[i], dp[j - 1] + 1)

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
const int MAX = 1000 + 10;
const int INF = 0x3f3f3f3f;
char s[MAX];
int dp[MAX],len,is_h[MAX][MAX];
void judge_huiwen()
{
    int x,y;
    memset(is_h, 0, sizeof(is_h));
    for(int i = 1; i <= len; i++)
    {
        x = i - 1;
        y = i + 1;
        while(x > 0 && y <= len && s[x] == s[y])
        {
            is_h[x][y] = 1;
            x--;
            y++;
        }
        x = i;
        y = i + 1;
        while(x > 0 && y <= len && s[x] == s[y])
        {
            is_h[x][y] = 1;
            x--;
            y++;
        }
    }
}
int main()
{
    int test;
    scanf("%d", &test);
    while(test--)
    {
        scanf("%s", s + 1);
        len = strlen(s + 1);
        memset(dp,0,sizeof(dp));
        dp[1] = 1;
        judge_huiwen();
        for(int i = 2; i <= len; i++)
        {
            dp[i] = dp[i - 1] + 1;
            for(int j = i - 1; j > 0; j--)
            {
                if(is_h[j][i])
                {
                     dp[i] = min(dp[i], dp[j - 1] + 1);
                }
            }
        }
        printf("%d\n",dp[len]);
    }
    return 0;
}