[NOIP 2016] 换教室

嘟嘟嘟

这道题很显然是dp，刚开始我是设dp[i][j]表示到第 i 个时间段，申请了 j 个课程耗费的体力值的总和的最小期望值，但是想了半天也没有推出转移方程，现在想想觉得应该是无法表示他第 i 时刻在哪一个教室。

于是我们加一维dp[i][j][0/1]表示第 i 时刻我们是否申请了换教室。那么dp方程就很好推出来了：

先令u1 = c[i - 1], v1 = c[i], u2 = d[i - 1], v2 = d[i]

dp[i][j][0] = min(dp[i - 1][j][0] + v[u1][v1], dp[i - 1][j][1] + v[u2][v1] * k[i - 1] + v[u1][v1] * (1 - k[i - 1]))

dp[i][j][1] = min(dp[i - 1][j - 1][0] + v[u1][v2] * k[i] + v[u1][v1] * (1 - k[i]),

　　　　　　　dp[i - 1][j - 1][1] + v[u2][v2] * k[i - 1] * k[i] + v[u1][v2] * (1 - k[i - 1]) * k[i] + v[u2][v1] * k[i - 1] * (1 - k[i]) + v[u1][v1] * (1 - k[i - 1]) * (1 - k[i]))

总的来说，就是如果这一次申请换教室的话，要同时考虑换教室成功和不成功的两种情况。

然后ans = min(dp[n][i][0], dp[n][i][1]) (0 <= i <= m,因为可以不申请换任何一个教室)

然后求多源最短路，又因为V <= 300,直接floyd预处理就行。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("") 
#define space putchar(' ')
#define Mem(a) memset(a, 0, sizeof(a))
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int max_nod = 305;
const int maxn = 2e3 + 5;
inline ll read()
{
    ll ans = 0;
    char ch = getchar(), last = ' ';
    while(!isdigit(ch)) {last = ch; ch = getchar();}
    while(isdigit(ch)) {ans = ans * 10 + ch - '0'; ch = getchar();}
    if(last == '-') ans = -ans;
    return ans;
}
inline void write(ll x)
{
    if(x < 0) x = -x, putchar('-');
    if(x >= 10) write(x / 10);
    putchar(x % 10 + '0');
}

int n, m, V, E;
ll v[max_nod][max_nod]; 
int a[maxn], b[maxn];
db p[maxn];
db dp[maxn][maxn][2];

void init()
{
    for(int i = 1; i <= V; ++i)
        for(int j = 1; j <= V; ++j) v[i][j] = INF;
    for(int i = 1; i <= n; ++i)
        for(int j = 0; j <= m; ++j) dp[i][j][0] = dp[i][j][1] = INF;        //j:0开始 
}

int main()
{
    n = read(); m = read(); V = read(); E = read();
    for(int i = 1; i <= n; ++i) a[i] = read();
    for(int i = 1; i <= n; ++i) b[i] = read();
    for(int i = 1; i <= n; ++i) scanf("%lf", &p[i]);
    init();
    for(int i = 1; i <= E; ++i)
    {
        int x = read(), y = read(); ll co = read();
        v[x][y] = v[y][x] = min(v[x][y], co);
    }
    for(int k = 1; k <= V; ++k)
        for(int i = 1; i <= V; ++i)
            for(int j = 1; j <= V; ++j)
                v[i][j] = min(v[i][j], v[i][k] + v[k][j]);
    for(int i = 1; i <= V; ++i) v[i][i] = v[i][0] = v[0][i] = 0;
    dp[1][0][0] = dp[1][1][1] = 0;
    for(int i = 2; i <= n; ++i)
    {
        dp[i][0][0] = dp[i - 1][0][0] + v[a[i - 1]][a[i]];
        for(int j = 1; j <= min(i, m); ++j)
        {
            db tp_0 = dp[i - 1][j][0] + v[a[i - 1]][a[i]];
            db tp_1 = dp[i - 1][j][1] + (db)v[b[i - 1]][a[i]] * p[i - 1] + (db)v[a[i - 1]][a[i]] * (1 - p[i - 1]);
            dp[i][j][0] = min(dp[i][j][0], min(tp_0, tp_1));
            tp_0 = dp[i - 1][j - 1][0] + (db)v[a[i - 1]][b[i]] * p[i] + (db)v[a[i - 1]][a[i]] * (1 - p[i]);
            tp_1 = dp[i - 1][j - 1][1] + (db)v[b[i - 1]][b[i]] * p[i - 1] * p[i] + (db)v[a[i - 1]][b[i]] * (1 - p[i - 1]) * p[i];
            tp_1 += (db)v[b[i - 1]][a[i]] * p[i - 1] * (1 - p[i]) + (db)v[a[i - 1]][a[i]] * (1 - p[i - 1]) * (1 - p[i]);
            dp[i][j][1] = min(dp[i][j][1], min(tp_0, tp_1));
        }
    }
    db ans = INF;
    for(int i = 0; i <= m; ++i) ans = min(ans, min(dp[n][i][0], dp[n][i][1]));
    printf("%.2lf
", ans);
    return 0;
}