14281: 【原4281】Palindrome partitioning

题目描述

author: Jarvis 原OJ链接：https://acm.sjtu.edu.cn/OnlineJudge-old/problem/4281

Description

Given a string s, partition s such that every substring of the partition is a palindrome.

Retrun the minimum number of cuts needed for a palindrome partitioning of s.

Input Format

One line. The string s.

Output Format

One line. The minimum number of cuts.

Sample Input

``````aab
``````

Sample Output

``````1
``````

Explanation

``````The palindrome partitioning ['aa', 'b'] could be produced using 1 cut.
``````

zqy2018's solution

``````/*
See the solution at https://github.com/zqy1018/sjtu_oj_solutions/blob/master/solutions/cs222quiz_2019.md
*/
#include <bits/stdc++.h>
#define INF 2000000000
using namespace std;
typedef long long ll;
int f = 1, x = 0;
char c = getchar();
while(c < '0' || c > '9'){if(c == '-') f = -f; c = getchar();}
while(c >= '0' && c <= '9')x = x * 10 + c - '0', c = getchar();
return f * x;
}
int n, f[3005];
char s[3005];
bool ok[3005][3005] = {0};
void init(){
scanf("%s", s);
n = strlen(s);
for (int i = 0; i < n; ++i){
ok[i][i] = true;
for (int j = i - 1, k = i + 1; j >= 0 && k < n && s[j] == s[k]; --j, ++k)
ok[j][k] = true;
for (int j = i, k = i + 1; j >= 0 && k < n && s[j] == s[k]; --j, ++k)
ok[j][k] = true;
}
}
void solve(){
memset(f, 0x3f, sizeof(f));
f[0] = 0;
for (int i = 0; i < n; ++i){
for (int j = 0; j <= i; ++j)
if (ok[j][i])
f[i + 1] = min(f[i + 1], f[j] + 1);
}
printf("%d\n", f[n] - 1);
}
int main(){
init();
solve();
return 0;
}
``````