codeforces 245H . Queries for Number of Palindromes 区间dp
给一个字符串， q个询问， 每次询问求出[l, r]里有多少个回文串。
区间dp， dp[l][r]表示[l, r]内有多少个回文串。 dp[l][r] = dp[l+1][r]+dp[l][r-1]-dp[l+1][r-1]+flag[l][r]， 如果是回文串flag[l][r]为1。
#include <iostream> #include <vector> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <map> #include <set> #include <string> #include <queue> #include <stack> #include <bitset> 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, n, a) for(int i = a; i<n; i++) #define fi first #define se second 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} }; string s; int dp[5005][5005], flag[5005][5005]; int judge(int l, int r) { if(~flag[l][r]) return flag[l][r]; if(l == r) return flag[l][r] = 1; if(l == r-1 &&s[l] == s[r]) return flag[l][r] = 1; int tmpl = l, tmpr = r; while(l<r) { if(s[l]!=s[r]) return flag[tmpl][tmpr] = 0; return flag[tmpl][tmpr] = judge(l+1, r-1); } } int dfs(int l, int r) { if(~dp[l][r]) return dp[l][r]; if(l>r) return dp[l][r] = 0; if(l == r) return dp[l][r] = 1; dp[l][r] = dfs(l+1, r)+dfs(l, r-1)-dfs(l+1, r-1)+judge(l, r); return dp[l][r]; } int main() { mem1(dp); mem1(flag); cin>>s; int n, a, b; cin>>n; dfs(0, s.size()-1); for(int i = 0; i<n; i++) { scanf("%d%d", &a, &b); printf("%d\n", dp[a-1][b-1]); } return 0; }
