51Nod - 1021 石子归并 区间DP入门-分析

题意：

合并相邻石子，使花费最小

思路：

相邻的这个条件，决定了是区间相关的最优问题，事实上这是个区间DP的入门题

区间DP 枚举区间长度，对于本区间获得最优解，当分析更大的区间时，可由这个区间推出

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<set>
#include<queue>
#include<stack>
#include<map>
#define PI acos(-1.0)
#define in freopen("in.txt", "r", stdin)
#define out freopen("out.txt", "w", stdout)

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 100 + 7, maxd = (1<<18)-1, mod = 1e9 + 7;
const int INF = 0x7f7f7f7f;

int n;
int a[maxn];
int sum[maxn] = {0};
int dp[maxn][maxn];

int main() {

scanf("%d", &n); sum[0] = 0;
memset(dp, INF, sizeof dp);
for(int i = 1; i <= n; ++i) {
scanf("%d", &a[i]);
sum[i] = sum[i-1] + a[i];
dp[i][i] = 0;
}
for(int len = 1; len <= n; ++len) {
for(int i = 1; i + len <= n; ++i) {
int j = i + len;
for(int k = i; k < j; ++k) {
dp[i][j] = min(dp[i][j], dp[i][k]+dp[k+1][j]+sum[j]-sum[i-1]);
}
}

}
cout << dp[1]
<< endl;

return 0;
}