Codeforces 596D Wilbur and Trees dp (看题解)

一直在考虑， 每一段的贡献， 没想到这个东西能直接dp。。因为所有的h都是一样的。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ull unsigned long long

using namespace std;

const int N =  + ;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + ;
const double eps = 1e-;
const double PI = acos(-);

int n, h, x[N], lft[N], rgt[N];
double ans, p, pr, pl;
double f[N][N][][];
bool vis[N][N][][];

double dp(int i, int j, int u, int v) {
    if(i > j) return ;
    if(vis[i][j][u][v]) return f[i][j][u][v];
    double& ans = f[i][j][u][v];
    double tmp1 = , tmp2 = ;
    if(!u) ans += 0.5 * pl * (dp(i + , j, , v) + min(h, x[i] - x[i - ])), tmp1 = min(h, x[i] - x[i - ]);
    else ans += 0.5 * pl * (dp(i + , j, , v) + min(h, x[i] - x[i - ] - h)), tmp1 = min(h, x[i] - x[i - ] - h);
    if(v) ans += 0.5 * pr * (dp(i, j - , u, ) + min(h, x[j + ] - x[j])), tmp2 = min(h, x[j + ] - x[j]);
    else ans += 0.5 * pr * (dp(i, j - , u, ) + min(h, x[j + ] - h - x[j])), tmp2 = min(h, x[j + ] - h - x[j]);
    int L = min(rgt[i], j), R = max(lft[j], i);
    ans += 0.5 * pr * (dp(L + , j, , v) + x[L] - x[i] + (L != j ? h : tmp2));
    ans += 0.5 * pl * (dp(i, R - , u, ) + x[j] - x[R] + (R != i ? h : tmp1));
    vis[i][j][u][v] = true;
    return ans;
}

int main() {
    scanf("%d%d%lf", &n, &h, &p);
    pl = p, pr =  - p;
    for(int i = ; i <= n; i++) scanf("%d", &x[i]);
    x[] = -inf, x[n + ] = inf;
    sort(x + , x +  + n);
    lft[] = , rgt[n] = n;
    for(int i = ; i <= n; i++)
        lft[i] = x[i] - h < x[i - ] ? lft[i - ] : i;
    for(int i = n - ; i >= ; i--)
        rgt[i] = x[i] + h > x[i + ] ? rgt[i + ] : i;
    printf("%.10f\n", dp(, n, , ));
    return ;
}

/*
*/