【NOIP2016】换教室 题解（期望DP）

前言：状态贼鸡儿多，眼睛快瞎了。

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题目链接

题目大意：给定$n(课程数),m(可换次数),v(教室数),e(无向边数)$，同时给定原定教室$c[i]$和可换教室$d[i]$，换教室成功概率为$k[i]$，边权为$w[i]$。问耗费体力的最小期望值。

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设$f[i][j][0/1]$表示上完$i$节课，换教室$j$次后（$0$表示此刻不换，$1$表示刺客换）的最小期望值。

$C1=c[i-1],C2=c[i],C3=d[i-1],C4=d[i],mp[i][j]表示i到j的距离。$

先考虑不换的情况：

f[i][j][]=min(f[i][j][],min(f[i-][j][]+dis[c[i-]][c[i]],f[i-][j][]+(-k[i-])*dis[c[i-]][c[i]]+k[i-]*dis[d[i-]][c[i]]))

考虑换的情况：

f[i][j][]=min(f[i][j][],min(f[i-][j-][]+dis[c[i-]][c[i]]*(-k[i])+dis[c[i-]][d[i]]*k[i],f[i-][j-][]+dis[d[i-]][d[i]]*k[i-]*k[i]+dis[d[i-]][c[i]]*k[i-]*(-k[i])+dis[c[i-]][c[i]]*(-k[i-])*(-k[i])+dis[c[i-]][d[i]]*(-k[i-])*k[i]))

预处理最短路可以用$Floyd$，$ans=min(ans,min(f[n][i][0],f[n][i][1]))$。

代码:

#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e3 + ;
const double inf = 1e17 + ;
int n, m, v, e, c[MAXN][], mp[][];
double k[MAXN], dp[MAXN][MAXN][], ans;
inline int read() {
    char ch = getchar(); int u = , f = ;
    while (!isdigit(ch)) {if (ch == '-')f = -; ch = getchar();}
    while (isdigit(ch)) {u = u *  + ch - ; ch = getchar();}return u * f;
}
int main(){
    memset(mp, , sizeof(mp));
    n = read(); m = read(); v = read(); e = read();
    for (register int i = ; i <= n; i++)c[i][] = read();
    for (register int i = ; i <= n; i++)c[i][] = read();
    for (register int i = ; i <= n; i++)scanf("%lf", &k[i]);
    for (register int i = ; i <= e; i++){
        int x = read(), y = read(), w = read();
        mp[x][y] = mp[y][x] = min(mp[x][y], w);
    }
    for (register int k = ; k <= v; k++)
        for (register int i = ; i <= v; i++)
            for (register int j = ; j <= v; j++)
                mp[i][j] = min(mp[i][j], mp[i][k] + mp[k][j]);
    for (register int i = ; i <= v; i++)mp[i][i] = mp[i][] = mp[][i] = ;
    for (register int i = ; i <= n; i++)
        for (register int j = ; j <= m; j++)dp[i][j][] = dp[i][j][] = inf;
    dp[][][] = dp[][][] = ;
    for (register int i = ; i <= n; i++){
        dp[i][][] = dp[i - ][][] + mp[c[i - ][]][c[i][]];
        for (register int j = ; j <= min(i, m); j++){
            int C1 = c[i - ][], C2 = c[i - ][], C3 = c[i][], C4 = c[i][];
            dp[i][j][] = min(dp[i][j][], min(dp[i - ][j][] + mp[C1][C3], dp[i - ][j][] + mp[C1][C3] * ( - k[i - ]) + mp[C2][C3] * k[i - ]));
            dp[i][j][] = min(dp[i][j][], min(dp[i - ][j - ][] + mp[C1][C3] * ( - k[i]) + mp[C1][C4] * k[i], dp[i - ][j - ][] + mp[C2][C4] * k[i] * k[i - ] + mp[C2][C3] * k[i - ] * ( - k[i]) + mp[C1][C4] * ( - k[i - ]) * k[i] + mp[C1][C3] * ( - k[i - ]) * ( - k[i])));
        }
    }
    ans = inf;
    for (register int i = ; i <= m; i++)ans = min(ans, min(dp[n][i][], dp[n][i][]));
    printf("%.2lf", ans);
    return ;
}