51Nod 1021 石子归并(区间dp经典入门)

题意：

N堆石子摆成一条线。现要将石子有次序地合并成一堆。规定每次只能选相邻的2堆石子合并成新的一堆，并将新的一堆石子数记为该次合并的代价。计算将N堆石子合并成一堆的最小代价。

n<=100

思路：

dp[i][j]表示以i开头，长度为j的石子合并的答案

dp[i][j] = min(dp[i][k] + dp[i+k][j-k], dp[i][j] + sum(i,i+j-1));

代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<vector>
#include<map>
#include<functional>
#include<list>

#define fst first
#define sc second
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lc root<<1
#define rc root<<1|1
#define lowbit(x) ((x)&(-x)) 

using namespace std;

typedef double db;
typedef long double ldb;
typedef long long ll;
typedef long long LL;
typedef unsigned long long ull;
typedef pair<int,int> PI;
typedef pair<ll,ll> PLL;

const db eps = 1e-;
const int mod = 1e9+;
const int maxn = 2e6+;
const int maxm = 2e6+;
const int inf = 0x3f3f3f3f;
//const db pi = acos(-1.0);
const int N = maxn;

int n;
int dp[][];
int a[maxn];
int s[maxn];
int main(){
    int n;
    scanf("%d", &n);
    mem(dp,inf);
    for(int i = ; i <= n; i++){
        scanf("%d", &a[i]);
        dp[i][] = ;
        s[i] = s[i-]+a[i];
    }
    for(int j = ; j <= n; j++){
        for(int i = ; i+j- <= n; i++){
            for(int k = ; k < j; k++){
                dp[i][j] = min(dp[i][j], dp[i][k] + dp[i+k][j-k]+s[i+j-]-s[i-]);
            }
            //printf("%d %d %d\n",i,j,dp[i][j]);
        }
    }
    printf("%d",dp[][n]);
    return ;
} 

/*`
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59 -17 5 67 87 50 -71 54 27 -10
 */