poj1651 区间dp

转移方程是dp[i][j]=min(dp[i][j],dp[i][k]+dp[k][j]+mul[i][k][j]);

k是中间的那个数,每次更新k的值

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<cstdio>
#include<cassert>
#include<iomanip>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define pii pair<int,int>
#define C 0.5772156649
#define pi acos(-1.0)
#define ll long long
#define mod 1000000007
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1

using namespace std;

const double g=10.0,eps=1e-7;
const int N=100+10,maxn=60000+10,inf=0x3f3f3f;

ll dp[N][N],a[N],mul[N][N][N];
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    ll n;
    cin>>n;
    for(int i=1;i<=n;i++)cin>>a[i];
    for(int i=1;i<=n;i++)
        for(int j=i+1;j<=n;j++)
            for(int k=j+1;k<=n;k++)
                mul[i][j][k]=a[i]*a[j]*a[k];
    for(int i=1;i<=n;i++)
    {
        for(int j=i+1;j<=n;j++)
        {
            if(i+1==j)dp[i][j]=0;
            else if(i+2==j)dp[i][j]=mul[i][i+1][i+2];
            else dp[i][j]=1e18;
        }
    }
    for(int len=2;len<=n;len++)
    {
        for(int i=1,j=len;j<=n;i++,j++)
        {
            for(int k=i+1;k<=j-1;k++)
            {
            //    cout<<dp[k][j]<<" "<<i<<" "<<k<<" "<<j<<endl;
                dp[i][j]=min(dp[i][j],dp[i][k]+dp[k][j]+mul[i][k][j]);
            }
        }
    }
    cout<<dp[1][n]<<endl;
    return 0;
}
/********************
4
10 1 50 5
********************/
View Code

 

posted @ 2017-09-14 10:12  walfy  阅读(149)  评论(0编辑  收藏  举报