UVa OJ 11584 - Partitioning by Palindromes

UVa OJ 11584 - Partitioning by Palindromes

Problem

Here is the: link

Solution

先对字符串进行预处理，把回文字符串的长度全部记录下来，然后用DP对回文字符串的个数进行处理。最小个数=min(之前已经处理过的长度所含回文字符串的最小值+未处理的长度所含回文字符数的最小值)

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

const int maxn = 1005;
char s[maxn];
int d[maxn], idx[maxn][maxn], cas;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
//freopen("input.txt" , "r", stdin );
//freopen("output.txt", "w", stdout);
cin >> cas;
while(cas--) {
cin >> s+1;
int n = strlen(s+1);
memset(d, 0x3f, sizeof(d));
d[0] = 0;
for (int i = 1; i <= n; ++i) {
idx[i][i] = 1;
for (int j = i+1; j <= n; ++j) {
bool isOK = false;
for (int k = 0; k < (j-i+1>>1); ++k)
if (s[i+k] != s[j-k]) {
idx[i][j] = j-i+1;
isOK = true;
break;
}
if (isOK) continue;
idx[i][j] = 1;
}
}
for (int i = 1; i <=n; ++i) {
for (int j = 0; j <=i; ++j) {
d[i] = min(d[i], d[j]+idx[j+1][i]);
}
}
cout << d
<< '\n';
}
return 0;
}