• bzoj4518


    http://www.lydsy.com/JudgeOnline/problem.php?id=4518

    水了一发斜率优化 好久没写了 调了一个晚上

    dp[i][j]:第i天走到了第j个休息站 

    dp[i][j] = min(dp[i-1][x] + (s[i]-s[x]-v)^2)v是平均数=s[n]/m

    然后我们化简一下得到了一个式子 (h[j]-h[k])/(s[j]-s[k])/2<=s[i]是不优的(这里我好像把小于等于号推反了,这里是对的)那么我们就可以斜率优化了

    然后写了很长时间

    注意:1.队列是l<r 2.要赋初值

    #include<bits/stdc++.h>
    using namespace std;
    typedef long long ll;
    const int N = 3010;
    int n, m;
    ll v;
    ll s[N], h[N], dp[N][N];
    int q[N];
    double slope(int j, int k) 
    {
        return ((double)h[j] - (double)h[k]) / ((double)s[j] - (double)s[k]) / 2.0;
    }
    void solve()
    {
        for(int i = 0; i <= n; ++i) dp[1][i] = (s[i] - v) * (s[i] - v);
        for(int i = 2; i <= m; ++i)
        {
            int l = 1, r = 0;  q[++r] = 0;
            dp[i][0] = v * v + dp[i - 1][0];
            for(int j = 0; j <= n; ++j) h[j] = dp[i - 1][j] + s[j] * s[j] + 2 * v * s[j];
            for(int j = 1; j <= n; ++j) 
            {
                while(l < r && slope(q[l], q[l + 1]) <= s[j]) ++l;
                dp[i][j] = min(dp[i - 1][q[l]] + (s[j] - s[q[l]] - v) * (s[j] - s[q[l]] - v), dp[i - 1][j] + v * v);
                while(l < r && slope(j, q[r]) <= slope(q[r - 1], q[r])) --r;
                q[++r] = j;
            }
        }
        printf("%lld
    ", dp[m][n] / m);
    }
    int main()
    {
        freopen("menci_journey.in", "r", stdin);
        freopen("menci_journey.out", "w", stdout);
        scanf("%d%d", &n, &m);
        for(int i = 1; i <= n; ++i)
        {
            ll x; scanf("%lld", &x);
            s[i] = s[i - 1] + x * (ll)m; 
        }
        v = s[n] / (ll)m;
        solve();
        fclose(stdin); fclose(stdout);
        return 0;
    }
    View Code
  • 相关阅读:
    google wave : email也能焕发新的青春
    twine语义网的平台
    www2009参会感受
    太激动了!一个计算机学科的YouTube!
    Ian Foster 网格和云计算
    搞笑学术论文chicken paper
    Web数据集汇总
    找不到符号System.out.printIn("hello world");
    jdk下载地址
    线程《一:创建线程的方式》
  • 原文地址:https://www.cnblogs.com/19992147orz/p/6746749.html
Copyright © 2020-2023  润新知