UVa OJ 11584 - Partitioning by Palindromes

Problem

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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] << '\n';    }    return 0;}