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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