UVa 11584 Partitioning by Palindromes (简单DP)

Posted 2020-08-27 dwtfukgv

题意：给定一个字符串，求出它最少可分成几个回文串。

析：dp[i] 表示前 i 个字符最少可分成几个回文串，dp[i] = min{ 1 + dp[j-1] | j-i是回文}。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <unordered_map>
#include <unordered_set>
#define debug() puts("++++");
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e3 + 5;
const int mod = 2000;
const int dr[] = {-1, 1, 0, 0};
const int dc[] = {0, 0, 1, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
char s[maxn];
int dp[maxn];

bool judge(int i, int j){
    while(i < j){
        if(s[i] != s[j])  return false;
        ++i, --j;
    }
    return true;
}

int main(){
    int T;  cin >> T;
    while(T--){
        scanf("%s", s);
        n = strlen(s);
        memset(dp, INF, sizeof dp);
        dp[0] = 1;
        for(int i = 1; i < n; ++i){
            for(int j = 0; j < i; ++j)
                if(judge(j, i)) dp[i] = min(dp[i], 1 + dp[j-1]);
                else dp[i] = min(dp[i], dp[i-1]+1);
        }
        printf("%d\n", dp[n-1]);

    }
    return 0;
}