给一个长度为n的序列， 每一次可以消去其中的一个回文串， 问最少几次才可以消完。
代码很清楚
#include <iostream> #include <vector> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <map> #include <set> #include <string> #include <queue> using namespace std; #define pb(x) push_back(x) #define ll long long #define mk(x, y) make_pair(x, y) #define lson l, m, rt<<1 #define mem(a) memset(a, 0, sizeof(a)) #define rson m+1, r, rt<<1|1 #define mem1(a) memset(a, -1, sizeof(a)) #define mem2(a) memset(a, 0x3f, sizeof(a)) #define rep(i, a, n) for(int i = a; i<n; i++) #define ull unsigned long long typedef pair<int, int> pll; const double PI = acos(-1.0); const double eps = 1e-8; const int mod = 1e9+7; const int inf = 1061109567; const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} }; int dp[505][505], a[505]; int dfs(int l, int r) { if(dp[l][r]) return dp[l][r]; dp[l][r] = inf; if(l == r) return dp[l][r] = 1; if(r-l == 1) return dp[l][r] = (a[l]==a[r])?1:2; if(a[l] == a[r]) dp[l][r] = dfs(l+1, r-1); for(int i = l; i<=r; i++) { dp[l][r] = min(dp[l][r], dfs(l, i)+dfs(i+1, r)); } return dp[l][r]; } int main() { int n; cin>>n; for(int i = 1; i<=n; i++) { scanf("%d", &a[i]); } cout<<dfs(1, n); return 0; }