HDU 5001 Walk(概率DP)
2014-09-15 01:26
405 查看
这题我一开始用DFS去写。会出现重复的问题。。后来看了别人递推的方法。其实好简单啊。还是做太少了,dp[i][j],代表第j步在i位置的概率,在推能到达某个位置的概率的时候,上一层要从不是要推的这个位置开始推下去,因为这一层如果到了,后面就不需要推了。
AC代码:
#include<cstdio>
#include<ctype.h>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<vector>
#include<cstdlib>
#include<stack>
#include<cmath>
#include<queue>
#include<set>
#include<map>
#include<ctime>
#include<string.h>
#include<string>
using namespace std;
#define ll __int64
#define eps 1e-10
#define MOD 10000007
double dp[55][10005];
vector<int>a[55];
int n,d;
double cal(int p)
{
int i,j,k;
double temp = 0;
for(i = 1; i <= d; i++)
{
for(j = 1; j <= n; j++)
{
if(j == p) continue;
int sz = a[j].size();
double db = 1.0/(double)sz;
for(k = 0; k < sz; k++)
dp[a[j][k]][i] += dp[j][i-1]*db;
}
temp += dp[p][i];
}
return temp;
}
int main()
{
#ifdef GLQ
freopen("input.txt","r",stdin);
// freopen("o2.txt","w",stdout);
#endif // GLQ
int t,i,j,m;
scanf("%d",&t);
while(t--)
{
scanf("%d%d%d",&n,&m,&d);
for(i = 1; i <= n; i++)
a[i].clear();
for(i = 1; i <= m; i++)
{
int t1,t2;
scanf("%d%d",&t1,&t2);
a[t1].push_back(t2);
a[t2].push_back(t1);
}
double fir = 1.0/(double)n,ans;
for(i = 1; i <= n; i++)
{
memset(dp,0,sizeof(dp));
for(j = 1; j <= n; j++)
dp[j][0] = fir;
ans = cal(i)+fir;
printf("%.10lf\n",1.0-ans);
}
}
return 0;
}
AC代码:
#include<cstdio>
#include<ctype.h>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<vector>
#include<cstdlib>
#include<stack>
#include<cmath>
#include<queue>
#include<set>
#include<map>
#include<ctime>
#include<string.h>
#include<string>
using namespace std;
#define ll __int64
#define eps 1e-10
#define MOD 10000007
double dp[55][10005];
vector<int>a[55];
int n,d;
double cal(int p)
{
int i,j,k;
double temp = 0;
for(i = 1; i <= d; i++)
{
for(j = 1; j <= n; j++)
{
if(j == p) continue;
int sz = a[j].size();
double db = 1.0/(double)sz;
for(k = 0; k < sz; k++)
dp[a[j][k]][i] += dp[j][i-1]*db;
}
temp += dp[p][i];
}
return temp;
}
int main()
{
#ifdef GLQ
freopen("input.txt","r",stdin);
// freopen("o2.txt","w",stdout);
#endif // GLQ
int t,i,j,m;
scanf("%d",&t);
while(t--)
{
scanf("%d%d%d",&n,&m,&d);
for(i = 1; i <= n; i++)
a[i].clear();
for(i = 1; i <= m; i++)
{
int t1,t2;
scanf("%d%d",&t1,&t2);
a[t1].push_back(t2);
a[t2].push_back(t1);
}
double fir = 1.0/(double)n,ans;
for(i = 1; i <= n; i++)
{
memset(dp,0,sizeof(dp));
for(j = 1; j <= n; j++)
dp[j][0] = fir;
ans = cal(i)+fir;
printf("%.10lf\n",1.0-ans);
}
}
return 0;
}
相关文章推荐
- hdu 5001 Walk 概率DP
- HDU 5001 Walk (暴力、概率dp)
- HDU 5001 Walk (概率dp)
- hdu 5001 Walk 【概率dp】
- 【HDU 5001】Walk(矩阵快速幂+概率DP)
- [概率dp] hdu 5001 Walk
- HDU-5001 Walk (概率DP)
- HDU 5001 Walk 概率DP
- hdu 5001 Walk (概率dp)
- hdu 5001 Walk 概率dp 2014 ACM/ICPC Asia Regional Anshan Online
- HDU 5001 Walk(暴力+概率DP)
- HDU 5001 Walk 概率DP BFS 矩阵递推 暴力
- HDU 5001 Walk 概率DP入门!
- [ACM] hdu 5001 Walk (概率DP)
- HDU 5001-Walk(概率dp)
- hdu 5001 Walk ( 概率DP )
- hdu5001 Walk 概率DP
- HDU - 5001 Walk(概率dp)
- HDU 5001 Walk (概率DP)
- hdu 5001 walk 概率dp入门题