hdu4035 Maze

题目链接

hdu4035 Maze

题解

f[u]表示在节点u通关的所需的边数期望

转移方程分叶子节点和非叶子点讨论

发现都可以化成f[x]=af[1]+bf[dad]+c的形式

然后推一下系数

还是看这个吧https://www.cnblogs.com/Paul-Guderian/p/7624039.html

代码

#include<cmath>
#include<cstdio>
#include<ctime>
#include<cstring>
#include<algorithm>
inline int read() {
    int x = 0,f = 1;
    char c = getchar();
    while(c < '0' || c > '9')c = getchar();
    while(c <= '9' && c >= '0') x = x * 10 + c - '0',c = getchar();
    return x * f;
}
#define eps 1e-9
const int maxn = 100007;
struct node {
    int v,next;
} edge[maxn << 1];
int head[maxn],num = 0;
inline void add_edge(int u,int v) {
    edge[++ num].v = v; edge[num].next = head[u]; head[u] = num;
}
int n;
int a[maxn];
double A[maxn],B[maxn],C[maxn];
double k[maxn],e[maxn];
bool dfs(int x,int fa) {
    if(!edge[head[x]].next && x != 1) {
        A[x] = k[x];
        B[x] = 1 - k[x] - e[x];
        C[x] = 1 - k[x] - e[x];
        return true;
    }
    double A_ = 0,B_ = 0,C_ = 0; int cnt = 0;
    for(int i = head[x];i;i = edge[i].next) {
        int v = edge[i].v;
        if(++ cnt && v != fa) {
            if(!dfs(v,x)) return false;
            A_ += A[v];B_ += B[v],C_ += C[v];
        }
    }
    if(fabs(1 - (1 - k[x] - e[x]) / cnt * B_) < eps) return false ;
    A[x] = (k[x] + (1 - k[x] - e[x]) / cnt * A_ ) / (1 - (1 - k[x] - e[x]) / cnt * B_);
    B[x] = ((1 - k[x] - e[x]) / cnt) / (1 - (1 - k[x] - e[x]) / cnt * B_);
    C[x] = (1 - k[x] - e[x] + (1 - k[x] - e[x]) / cnt * C_) / (1 - (1 - k[x] - e[x]) / cnt * B_);
    return true;
}
int main() {
    int Q = read();
    for(int t = 1;t <= Q;t += 1) {
        memset(head,0,sizeof head);
        num = 0;
        n = read();
        for(int i = 1;i < n;++ i) {
            int u = read(),v = read();
            add_edge(u,v);add_edge(v,u);
        }
        for(int i = 1;i <= n;++ i) {
            scanf("%lf%lf",&k[i],&e[i]);
            k[i] /= 100,e[i] /= 100;
        }
        printf("Case %d : ",t);
        if(!dfs(1,1) || fabs(1 - A[1]) < eps) {
            puts("impossible"); continue;
        } else printf("%.6lf\n",C[1] / (1 - A[1]));
    }
}