紫书动规 例题9-7 UVA - 11584 Partitioning by Palindromes dp

题目链接:

https://vjudge.net/problem/UVA-11584

题意:

题解:

dp[i]:=考虑到第i个字符，的最少划分
dp[i] = min{dp[j]+1} (0<=j

代码:

#include 
using namespace std;
typedef long long ll;
#define MS(a) memset(a,0,sizeof(a))
#define MP make_pair
#define PB push_back
const int INF = 0x3f3f3f3f;
const ll INFLL = 0x3f3f3f3f3f3f3f3fLL;
inline ll read(){
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
//////////////////////////////////////////////////////////////////////////
const int maxn = 1e3+10;

int n,kase,vis[maxn][maxn],p[maxn][maxn],dp[maxn];
char s[maxn];

int is_palindrome(int i,int j){
    if(i>j) return 1;
    if(s[i] != s[j]) return 0;

    if(vis[i][j] == kase) return p[i][j];
    vis[i][j] = kase;
    p[i][j] = is_palindrome(i+1,j-1);
    return p[i][j];
}

int main(){
    int T = read();
    for(kase=1; kase<=T; kase++){
        scanf("%s",s+1);
        // dp[i] = min(dp[i],dp[j]+1); is_palindrome(j+1,i);
        int len = strlen(s+1);
        // memset(dp,0x3f,sizeof(dp));
        dp[0] = 0;
        for(int i=1; i<=len; i++){
            dp[i] = dp[i-1]+1;
            for(int j=0; jif(is_palindrome(j+1,i))
                    dp[i] = min(dp[i],dp[j]+1);
                // cout << i << " " << dp[i] << " \n";
            }
        }

        cout << dp[len] << endl;
    }

    return 0;
}

// 3
// racecar
// fastcar
// aaadbccb