UVA 11584 Partitioning by Palindromes——dp

n^2预处理一下a【i】【j】是否为回文串，处理的方法比较笨，分奇偶写的。。。

剩下的dp思路很好想，不多说了

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 1010;
const int INF = 0x3f3f3f3f;
int T, n;
char s[maxn];
bool a[maxn][maxn];
int dp[maxn];
int main() {
    scanf("%d", &T);
    getchar();
    while (T--) {
        gets(s+1);
        n = strlen(s+1);
        memset(a, false, sizeof(a));
        for (int i = 1; i <= n; i++) a[i][i] = true;
        for (int i = 2; i < n; i++) {
            int x = i - 1, y = i + 1;
            while (x >= 1 && y <= n && s[x] == s[y]) {
                a[x][y] = true;
                x--; y++;
            }
        }
        for (int i = 1; i <= n; i++) {
            int x = i, y = i+1;
            while (x >= 1 && y <= n && s[x] == s[y]) {
                a[x][y] = true;
                x--; y++;
            }
        }
        dp[0] = 0;
        for (int i = 1; i <= n; i++) {
            dp[i] = i;
            for (int j = 0; j < i; j++) {
                if (a[j+1][i]) dp[i] = min(dp[i], dp[j] + 1);
            }
        }
        printf("%d\n", dp[n]);
    }
    return 0;
}