http://lx.lanqiao.cn/problem.page?gpid=T414
题意:……
思路:很普通的区间DP,但是因为n<=1000,所以O(n^3)只能拿90分。上网查了下了解了平行四边形优化:地址。
但是看不懂。
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long LL; 4 const LL INF = 100000000000000000LL; 5 LL dp[1010][1010], s[1010][1010]; 6 LL sum[1010]; 7 int main() { 8 LL ans = 0; 9 int n; scanf("%d", &n); 10 for(int i = 1; i <= n; i++) scanf("%lld", &dp[i][i]), sum[i] = sum[i-1] + dp[i][i], dp[i][i] = 0, s[i][i] = i; 11 for(int len = 1; len < n; len++) { 12 for(int l = 1; l + len <= n; l++) { 13 int r = l + len; 14 dp[l][r] = INF; 15 for(int k = s[l][r-1]; k <= s[l+1][r]; k++) { 16 LL now = dp[l][k] + dp[k+1][r] + sum[r] - sum[l - 1]; 17 if(dp[l][r] > now) dp[l][r] = now, s[l][r] = k; 18 } 19 } 20 } 21 printf("%lld ", dp[1][n]); 22 return 0; 23 }