Codeforces 596D Wilbur and Trees dp (看题解)

原文链接： http://www.cnblogs.com/CJLHY/p/10512296.html

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 = 2000 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);

int n, h, x[N], lft[N], rgt[N];
double ans, p, pr, pl;
double f[N][N][2][2];
bool vis[N][N][2][2];

double dp(int i, int j, int u, int v) {
    if(i > j) return 0;
    if(vis[i][j][u][v]) return f[i][j][u][v];
    double& ans = f[i][j][u][v];
    double tmp1 = 0, tmp2 = 0;
    if(!u) ans += 0.5 * pl * (dp(i + 1, j, 0, v) + min(h, x[i] - x[i - 1])), tmp1 = min(h, x[i] - x[i - 1]);
    else ans += 0.5 * pl * (dp(i + 1, j, 0, v) + min(h, x[i] - x[i - 1] - h)), tmp1 = min(h, x[i] - x[i - 1] - h);
    if(v) ans += 0.5 * pr * (dp(i, j - 1, u, 1) + min(h, x[j + 1] - x[j])), tmp2 = min(h, x[j + 1] - x[j]);
    else ans += 0.5 * pr * (dp(i, j - 1, u, 1) + min(h, x[j + 1] - h - x[j])), tmp2 = min(h, x[j + 1] - h - x[j]);
    int L = min(rgt[i], j), R = max(lft[j], i);
    ans += 0.5 * pr * (dp(L + 1, j, 1, v) + x[L] - x[i] + (L != j ? h : tmp2));
    ans += 0.5 * pl * (dp(i, R - 1, u, 0) + 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 = 1 - p;
    for(int i = 1; i <= n; i++) scanf("%d", &x[i]);
    x[0] = -inf, x[n + 1] = inf;
    sort(x + 1, x + 1 + n);
    lft[1] = 1, rgt[n] = n;
    for(int i = 2; i <= n; i++)
        lft[i] = x[i] - h < x[i - 1] ? lft[i - 1] : i;
    for(int i = n - 1; i >= 1; i--)
        rgt[i] = x[i] + h > x[i + 1] ? rgt[i + 1] : i;
    printf("%.10f\n", dp(1, n, 0, 1));
    return 0;
}

/*
*/

转载于:https://www.cnblogs.com/CJLHY/p/10512296.html