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题目链接:http://acm.swust.edu.cn/problem/code/745255/
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1 #include <iostream> 2 #include <cstring> 3 using namespace std; 4 #define min(a,b) a<b?a:b 5 #define inf 0x3f3f3f3f 6 7 int sum[201], dp[201][201]; //dp[i][j]表示区间[i,j]的最小代价 8 int main() 9 { 10 int n, i, j, k, r, x; 11 cin >> n; 12 memset(dp, inf, sizeof(dp)); 13 for (i = 1; i <= n; i++){ 14 cin >> x; 15 sum[i] = sum[i - 1] + x; 16 dp[i][i] = 0; 17 } 18 for (r = 0; r < n; r++){ 19 for (i = 1; i <= n - r; i++){ 20 j = i + r; 21 for (k = i; k < j; k++){ 22 dp[i][j] = min(dp[i][j], dp[i][k] + dp[k + 1][j] + sum[j] - sum[i - 1]); 23 } 24 } 25 } 26 cout << dp[1][n] << endl;; 27 return 0; 28 }
用类似的dp方法可以解决[Swust OJ 574]RentBoat [Swust oj 419]括号配对 [nyoj 37]回文字符串
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原文地址:http://www.cnblogs.com/zyxStar/p/4585591.html