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  • POJ 1260

    //状态转移方程:   F[i] = min{f[k] + (a[k+1]+………+a[i]+10} * p[i]}
    #include <iostream>
    #define MAXN 105
    using namespace std;
    
    int dp[MAXN];
    int n[MAXN];
    int p[MAXN];
    
    int main()
    {
        //freopen("acm.acm","r",stdin);
        int test;
        int c;
        int sum;
        int tem;
        int i;
        int j;
        int k;
        int min;
        int cur;
        cin>>test;
    
        while(test --)
        {
            memset(dp,0,sizeof(dp));
            cin>>c;
            for(i = 0; i < c; ++ i)
            {
                cin>>n[i];
                cin>>p[i];
            }
            sum = 0;
            
            
            dp[0] = (n[0]+10)*p[0];
            sum += n[0];
            for(i = 1; i < c; ++ i)
            {
                sum += n[i];
                cur = sum;
                min = 123456789;
                tem = (cur+10)*p[i];
                if(tem < min)
                {
                    min = tem;
                }
                for(k = 0; k < i; ++ k)
                {
                    cur -= n[k];
                    tem = dp[k] + (cur+10)*p[i];
                    if(tem < min)
                    {
                        min = tem;
                    }
    
                }
                dp[i] = min;
            }
            cout<<dp[c-1]<<endl;
        }
    }

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  • 原文地址:https://www.cnblogs.com/gavinsp/p/4563349.html
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