HDU4035 Maze

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神一般的概率DP

基本的状态转移方程：

dp[i] = k[i]*dp[1] + e[i]*0 + (1-k[i]-e[i])/d[i]* ( ∑( dp[j]+1 ); (i, j相连)

把原图看成一棵树，可以把每一项的状态转移方程化简成只与其父节点有关。

然后写出来的话会比较复杂= =

我说的可能不够清楚，详见kuangbin大神的http://www.cnblogs.com/kuangbin/archive/2012/10/03/2711108.html

怒贴源代码：

#include <cstdio>#include <cstring>#include <cstdlib>#include <ctime>#include <climits>#include <cmath>#include <iostream>#include <string>#include <vector>#include <set>#include <map>#include <list>#include <queue>#include <stack>#include <deque>#include <algorithm>using namespace std;#define ll long long#define pii pair<int, int>#define fr first#define sc second#define mp make_pair#define FOR(i,j,k) for(int i=j;i<=(k);i++)#define FORD(i,j,k) for(int i=j;i>=(k);i--)#define REP(i,n) for(int i=0;i<(n);i++)const double eps = 1e-9;const int maxn = 10005;int T, cas = 1, n;double k[maxn], e[maxn];double dp[maxn], a[maxn], b[maxn], c[maxn];vector<int> g[maxn];bool dfs(int u, int fa){    double sa = 0, sb = 0, sc = 0;    for (int i=0;i<g[u].size();i++)    {        int v = g[u][i];        if (v == fa) continue;        if (!dfs(v, u)) return 0;        sa += a[v]; sb += b[v]; sc += c[v];    }    sa *= (1-k[u]-e[u]) / g[u].size();    sb *= (1-k[u]-e[u]) / g[u].size();    sc *= (1-k[u]-e[u]) / g[u].size();    if (fabs(sb - 1) < eps) return 0;    a[u] = (k[u] + sa) / (1 - sb);    b[u] = (1-k[u]-e[u]) / g[u].size() / (1 - sb);    c[u] = (1-k[u]-e[u] + sc) / (1 - sb);    return 1;}int main(){    scanf("%d", &T);    while (T--)    {        scanf("%d", &n);        for (int i=0;i<=n;i++) g[i].clear();        for (int i=1,u,v;i<n;i++)        {            scanf("%d%d", &u, &v);            g[u].push_back(v);            g[v].push_back(u);        }        for (int i=1;i<=n;i++)        {            scanf("%lf%lf", &k[i], &e[i]);            k[i] /= 100; e[i] /= 100;        }        printf("Case %d: ", cas++);        if (!dfs(1, 0) || fabs(a[1] - 1) < eps) puts("impossible");        else printf("%.6lf\n", c[1] / (1 - a[1]));    }return 0;}