HDU4035 Maze(期望DP)

题意

抄袭自https://www.cnblogs.com/Paul-Guderian/p/7624039.html

有n个房间，由n-1条隧道连通起来，形成一棵树，从结点1出发，开始走，在每个结点i都有3种可能（概率之和为1）：1.被杀死，回到结点1处（概率为ki）2.找到出口，走出迷宫 （概率为ei）
3.和该点相连有m条边，随机走一条求：走出迷宫所要走的边数的期望值。(2≤n≤10000)

Sol

非常nice的一道题。

我简单的说一下思路：首先列出方程，$f[i]$表示在第$i$个位置走出迷宫的期望步数。

转移方程分叶子节点和父亲节点讨论一下，发现都可以化成$f[x] = a f[1] + b f[fa] + c$的形式

然后直接递推系数即可

具体可以看https://www.cnblogs.com/Paul-Guderian/p/7624039.html

/*

*/
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<set>
#include<queue>
#include<cmath>
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
//#define int long long
//#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1<<22, stdin), p1 == p2) ? EOF : *p1++)
//char buf[(1 << 22)], *p1 = buf, *p2 = buf;
using namespace std;
const int MAXN = 1e5 + 10, INF = 1e9 + 10;
const double eps = 1e-10;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < ‘0‘ || c > ‘9‘) {if(c == ‘-‘) f = -1; c = getchar();}
    while(c >= ‘0‘ && c <= ‘9‘) x = x * 10 + c - ‘0‘, c = getchar();
    return x * f;
}
int N;
vector<int> v[MAXN];
double b[MAXN], e[MAXN], A[MAXN], B[MAXN], C[MAXN];
bool dcmp(double x) {
    if(fabs(x) < eps) return 0;
    else return 1;
}
void init() {
    for(int i = 1; i <= N; i++) v[i].clear();
}
double Get(int x) {
    return (1 - b[x] - e[x]) / (v[x].size());
}
bool dfs(int x, int fa) {
    if(v[x].size() == 1 && (v[x][0] == fa)) {A[x] = b[x], C[x] = B[x] = Get(x); return 1;}
    double As = 0, Bs = 0, Cs = 0;
    for(int i = 0; i < v[x].size(); i++) {
        int to = v[x][i];
        if(to == fa) continue;
        if(!dfs(to, x)) return 0;
        As += A[to]; Bs += B[to]; Cs += C[to] + 1;
    }
    double P = Get(x);
    double D = (1 - Bs * P);
    if(!dcmp(D)) return 0;
    A[x] = (b[x] + As * P) / D;
    B[x] = P / D;
    C[x] = (Cs * P + ((x == 1) ? 0 : P)) / D;
    return 1;
}
int main() {
    int T = read();
    for(int GG = 1; GG <= T; GG++) {
        N = read(); init();
        //printf("%d ", v[3].size());
        for(int i = 1; i <= N - 1; i++) {
            int x = read(), y = read();
            v[x].push_back(y); v[y].push_back(x);
        }
        for(int i = 1; i <= N; i++) b[i] = (double) read() / 100, e[i] = (double) read() / 100;
        if(dfs(1, 0) && (dcmp(1 - A[1]))) printf("Case %d: %.10lf\n", GG, C[1] / (1 - A[1]));
        else printf("Case %d: impossible\n", GG);
    }
    return 0;
}
/*

*/

原文地址：https://www.cnblogs.com/zwfymqz/p/9527740.html