P1850 [NOIP2016 提高组] 换教室

最新推荐文章于 2025-01-02 14:49:38 发布

luyiming123

最新推荐文章于 2025-01-02 14:49:38 发布

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文章标签：图论

本文链接：https://blog.youkuaiyun.com/woshidalaocxy/article/details/133074332

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该博客主要探讨了一个利用动态规划和最短路径算法解决课程申请问题的解决方案。通过预处理教室之间的最短距离，并建立状态转移方程，博主详细解释了如何计算在有限课程选择下，使得教室切换成本期望最小的策略。博客内容涉及图论、动态规划和概率计算，适合对算法和优化问题感兴趣的读者阅读。

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Problem

Solution

设\(dp[i][j][0/1]\)表示在前\(i\)节课中，申请\(j\)节，同时第\(i\)节课申不申请的期望值。
首先需要Floyd预处理出各个教室之间最短路。
不难发现\(dp[i][..][..]\)一定由\(dp[i - 1][..][..]\)转移来。
分类讨论：

\(i\)不申请
- \(i - 1\)不申请\(dp[i][j][0] = dp[i - 1][j][0] + w[c[i - 1]][c[i]]\)
- \(i - 1\)申请\(dp[i][j][0] = dp[i - 1][j][1] + w[d[i - 1]][c[i]] \cdot k[i - 1] + w[c[i - 1]][c[i]] \cdot (1 - k[i - 1])\)
\(i\)申请
- \(i - 1\)不申请\(dp[i][j][1] = dp[i - 1][j - 1][0] + w[c[i - 1]][d[i]] \cdot k[i] + w[c[i - 1]][c[i]] \cdot (1 - k[i])\)
- \(i - 1\)申请\(dp[i][j][1] = dp[i - 1][j - 1][1] + w[d[i - 1]][d[i]] \cdot k[i - 1] \cdot k[i] + w[d[i - 1]][c[i]] \cdot k[i - 1] \cdot (1 - k[i]) + w[c[i - 1]][d[i]] \cdot (1 - k[i - 1]) \cdot k[i] + w[c[i - 1]][c[i]] \cdot (1 - k[i - 1]) \cdot (1 - k[i])\)

# include <bits/stdc++.h>
using namespace std;
int n,m,V,E;
int w[305][305];
int c[2005],d[2005];
double K[2005];
double dp[2005][2005][2];
const int inf = 0x3f3f3f3f;
int main(void)
{
    // freopen("P1850_2.in","r",stdin);
    scanf("%d%d%d%d",&n,&m,&V,&E);
    for(int i = 1; i <= n; i++) scanf("%d",&c[i]);
    for(int i = 1; i <= n; i++) scanf("%d",&d[i]);
    for(int i = 1; i <= n; i++) scanf("%lf",&K[i]);
    memset(w,0x3f3f3f,sizeof(w));
    for(int i = 1; i <= V; i++) w[i][i] = 0;
    for(int i = 1; i <= E; i++)
    {
        int a,b,_w;
        scanf("%d%d%d",&a,&b,&_w);
        w[a][b] = w[b][a] = min(w[a][b],_w);
    }
    for(int k = 1; k <= V; k++)
    {
        for(int i = 1; i <= V; i++)
        {
            for(int j = 1; j <= V; j++) 
            {
                if(i == j) continue;
                w[i][j] = min(w[i][j],w[i][k] + w[k][j]);
            }
        }
    }
    for(int i = 1; i <= n; i++)
    {
        for(int j = 0; j <= m; j++)
        {
            dp[i][j][0] = dp[i][j][1] = (double)inf;
        }
    }
    dp[1][0][0] = dp[1][1][1] = 0;
    for(int i = 2; i <= n; i++)
    {
        for(int j = 0; j <= min(i,m); j++)
        {
            dp[i][j][0] = min(dp[i][j][0],dp[i - 1][j][0] + (double)w[c[i - 1]][c[i]]);
            dp[i][j][0] = min(dp[i][j][0],dp[i - 1][j][1] + (double)w[d[i - 1]][c[i]] * K[i - 1] + (double)w[c[i - 1]][c[i]] * (1 - K[i - 1]));
            if(j == 0) continue;
            dp[i][j][1] = min(dp[i][j][1],dp[i - 1][j - 1][0] + (double)w[c[i - 1]][d[i]] * K[i] + (double)w[c[i - 1]][c[i]] * (1 - K[i]));
            dp[i][j][1] = min(dp[i][j][1],dp[i - 1][j - 1][1] + (double)w[d[i - 1]][d[i]] * K[i - 1] * K[i] + (double)w[d[i - 1]][c[i]] * K[i - 1] * (1 - K[i]) + (double)w[c[i - 1]][d[i]] * (1 - K[i - 1]) * K[i] + (double)w[c[i - 1]][c[i]] * (1 - K[i - 1]) * (1 - K[i]));
        }
    }
    double ans = (double)inf;
    for(int i = 0; i <= m; i++)
    {
        for(int j = 0; j <= 1; j++)
        {
            if(j == 1 && i == 0) continue;
            ans = min(ans,dp[n][i][j]);
        }
    }
    printf("%.2lf\n",ans);
    return 0;
}