Hdu 4035 Maze (dp求期望) - 2011 ACM/ICPC 成都赛区网络预选赛 1005
2011-09-15 09:30
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比赛时看题了,但是没有思路。比赛结束后这题总共通过20+,赛后看这个解题报告,由于博主说得太简洁,而我又是从来没有见过这种dp求数学期望的题,所以研究了好久都木有明白。只有搜索一下【dp求期望】的题目,从简单的开始入手,费了老大功夫,终于搞懂了,于是写下详细解题报告。
如果感觉这个题看不懂,也可以按照我的步骤来看:Poj 2096 -->
Zoj 3329 -->
Hdu 4035
/**
dp求期望的题。
题意:
有n个房间,由n-1条隧道连通起来,实际上就形成了一棵树,
从结点1出发,开始走,在每个结点i都有3种可能:
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。
设 E[i]表示在结点i处,要走出迷宫所要走的边数的期望。E[1]即为所求。
叶子结点:
E[i] = ki*E[1] + ei*0 + (1-ki-ei)*(E[father[i]] + 1);
= ki*E[1] + (1-ki-ei)*E[father[i]] + (1-ki-ei);
非叶子结点:(m为与结点相连的边数)
E[i] = ki*E[1] + ei*0 + (1-ki-ei)/m*( E[father[i]]+1 + ∑( E[child[i]]+1 ) );
= ki*E[1] + (1-ki-ei)/m*E[father[i]] + (1-ki-ei)/m*∑(E[child[i]]) + (1-ki-ei);
设对每个结点:E[i] = Ai*E[1] + Bi*E[father[i]] + Ci;
对于非叶子结点i,设j为i的孩子结点,则
∑(E[child[i]]) = ∑E[j]
= ∑(Aj*E[1] + Bj*E[father[j]] + Cj)
= ∑(Aj*E[1] + Bj*E[i] + Cj)
带入上面的式子得
(1 - (1-ki-ei)/m*∑Bj)*E[i] = (ki+(1-ki-ei)/m*∑Aj)*E[1] + (1-ki-ei)/m*E[father[i]] + (1-ki-ei) + (1-ki-ei)/m*∑Cj;
由此可得
Ai = (ki+(1-ki-ei)/m*∑Aj) / (1 - (1-ki-ei)/m*∑Bj);
Bi = (1-ki-ei)/m / (1 - (1-ki-ei)/m*∑Bj);
Ci = ( (1-ki-ei)+(1-ki-ei)/m*∑Cj ) / (1 - (1-ki-ei)/m*∑Bj);
对于叶子结点
Ai = ki;
Bi = 1 - ki - ei;
Ci = 1 - ki - ei;
从叶子结点开始,直到算出 A1,B1,C1;
E[1] = A1*E[1] + B1*0 + C1;
所以
E[1] = C1 / (1 - A1);
若 A1趋近于1则无解...
**/
#include <cstdio>
#include <iostream>
#include <vector>
#include <cmath>
using namespace std;
const int MAXN = 10000 + 5;
double e[MAXN], k[MAXN];
double A[MAXN], B[MAXN], C[MAXN];
vector<int> v[MAXN];
bool search(int i, int fa)
{
if ( v[i].size() == 1 && fa != -1 )
{
A[i] = k[i];
B[i] = 1 - k[i] - e[i];
C[i] = 1 - k[i] - e[i];
return true;
}
A[i] = k[i];
B[i] = (1 - k[i] - e[i]) / v[i].size();
C[i] = 1 - k[i] - e[i];
double tmp = 0;
for (int j = 0; j < (int)v[i].size(); j++)
{
if ( v[i][j] == fa ) continue;
if ( !search(v[i][j], i) ) return false;
A[i] += A[v[i][j]] * B[i];
C[i] += C[v[i][j]] * B[i];
tmp += B[v[i][j]] * B[i];
}
if ( fabs(tmp - 1) < 1e-10 ) return false;
A[i] /= 1 - tmp;
B[i] /= 1 - tmp;
C[i] /= 1 - tmp;
return true;
}
int main()
{
int nc, n, s, t;
cin >> nc;
for (int ca = 1; ca <= nc; ca++)
{
cin >> n;
for (int i = 1; i <= n; i++)
v[i].clear();
for (int i = 1; i < n; i++)
{
cin >> s >> t;
v[s].push_back(t);
v[t].push_back(s);
}
for (int i = 1; i <= n; i++)
{
cin >> k[i] >> e[i];
k[i] /= 100.0;
e[i] /= 100.0;
}
cout << "Case " << ca << ": ";
if ( search(1, -1) && fabs(1 - A[1]) > 1e-10 )
cout << C[1]/(1 - A[1]) << endl;
else
cout << "impossible" << endl;
}
return 0;
}
如果感觉这个题看不懂,也可以按照我的步骤来看:Poj 2096 -->
Zoj 3329 -->
Hdu 4035
/**
dp求期望的题。
题意:
有n个房间,由n-1条隧道连通起来,实际上就形成了一棵树,
从结点1出发,开始走,在每个结点i都有3种可能:
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。
设 E[i]表示在结点i处,要走出迷宫所要走的边数的期望。E[1]即为所求。
叶子结点:
E[i] = ki*E[1] + ei*0 + (1-ki-ei)*(E[father[i]] + 1);
= ki*E[1] + (1-ki-ei)*E[father[i]] + (1-ki-ei);
非叶子结点:(m为与结点相连的边数)
E[i] = ki*E[1] + ei*0 + (1-ki-ei)/m*( E[father[i]]+1 + ∑( E[child[i]]+1 ) );
= ki*E[1] + (1-ki-ei)/m*E[father[i]] + (1-ki-ei)/m*∑(E[child[i]]) + (1-ki-ei);
设对每个结点:E[i] = Ai*E[1] + Bi*E[father[i]] + Ci;
对于非叶子结点i,设j为i的孩子结点,则
∑(E[child[i]]) = ∑E[j]
= ∑(Aj*E[1] + Bj*E[father[j]] + Cj)
= ∑(Aj*E[1] + Bj*E[i] + Cj)
带入上面的式子得
(1 - (1-ki-ei)/m*∑Bj)*E[i] = (ki+(1-ki-ei)/m*∑Aj)*E[1] + (1-ki-ei)/m*E[father[i]] + (1-ki-ei) + (1-ki-ei)/m*∑Cj;
由此可得
Ai = (ki+(1-ki-ei)/m*∑Aj) / (1 - (1-ki-ei)/m*∑Bj);
Bi = (1-ki-ei)/m / (1 - (1-ki-ei)/m*∑Bj);
Ci = ( (1-ki-ei)+(1-ki-ei)/m*∑Cj ) / (1 - (1-ki-ei)/m*∑Bj);
对于叶子结点
Ai = ki;
Bi = 1 - ki - ei;
Ci = 1 - ki - ei;
从叶子结点开始,直到算出 A1,B1,C1;
E[1] = A1*E[1] + B1*0 + C1;
所以
E[1] = C1 / (1 - A1);
若 A1趋近于1则无解...
**/
#include <cstdio>
#include <iostream>
#include <vector>
#include <cmath>
using namespace std;
const int MAXN = 10000 + 5;
double e[MAXN], k[MAXN];
double A[MAXN], B[MAXN], C[MAXN];
vector<int> v[MAXN];
bool search(int i, int fa)
{
if ( v[i].size() == 1 && fa != -1 )
{
A[i] = k[i];
B[i] = 1 - k[i] - e[i];
C[i] = 1 - k[i] - e[i];
return true;
}
A[i] = k[i];
B[i] = (1 - k[i] - e[i]) / v[i].size();
C[i] = 1 - k[i] - e[i];
double tmp = 0;
for (int j = 0; j < (int)v[i].size(); j++)
{
if ( v[i][j] == fa ) continue;
if ( !search(v[i][j], i) ) return false;
A[i] += A[v[i][j]] * B[i];
C[i] += C[v[i][j]] * B[i];
tmp += B[v[i][j]] * B[i];
}
if ( fabs(tmp - 1) < 1e-10 ) return false;
A[i] /= 1 - tmp;
B[i] /= 1 - tmp;
C[i] /= 1 - tmp;
return true;
}
int main()
{
int nc, n, s, t;
cin >> nc;
for (int ca = 1; ca <= nc; ca++)
{
cin >> n;
for (int i = 1; i <= n; i++)
v[i].clear();
for (int i = 1; i < n; i++)
{
cin >> s >> t;
v[s].push_back(t);
v[t].push_back(s);
}
for (int i = 1; i <= n; i++)
{
cin >> k[i] >> e[i];
k[i] /= 100.0;
e[i] /= 100.0;
}
cout << "Case " << ca << ": ";
if ( search(1, -1) && fabs(1 - A[1]) > 1e-10 )
cout << C[1]/(1 - A[1]) << endl;
else
cout << "impossible" << endl;
}
return 0;
}
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