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Description I used to think I could be anything, but now I know that I couldn‘t do anything. So I started traveling. Input The first line contains an integer T, denoting the number of the test cases. Output For each test cases, output n lines, the i-th line containing the desired probability for the i-th node. Sample Input 2 5 10 100 1 2 2 3 3 4 4 5 1 5 2 4 3 5 2 5 1 4 1 3 10 10 10 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 4 9 Sample Output 0.0000000000 0.0000000000 0.0000000000 0.0000000000 0.0000000000 0.6993317967 0.5864284952 0.4440860821 0.2275896991 0.4294074591 0.4851048742 0.4896018842 0.4525044250 0.3406567483 0.6421630037 |
题目大意：在一幅图中，主人公随机选择一个点为起始点，然后每一步都是随机选择相邻点进行访问，一共走了d步。求各个点没有被访问的概率。
概率dp入门题，定义dp[u][i][j]：经过了j步之后落在i点且从来不经过u点的概率。最后对于各个点，从不经过它的概率就是dp[u][*][d]的和。
在实际操作过程中，可以把dp数组变成二维的，即变成dp[i][j]。
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
using namespace std;
const int maxn = 55;
int head[maxn],cur;
struct Edge
{
int to;
int next;
} edge[maxn*2];
void init()
{
cur=0;
memset(head,-1,sizeof(head));
}
void addedge(int u,int v)
{
edge[cur].to = v;
edge[cur].next = head[u];
head[u] = cur++;
}
double dp[maxn][10005];
double ans[maxn];
int du[maxn],n;
void cal_dp(int x,int d)
{
for(int u=1; u<=n; u++)
{
dp[u][d] = 0;
if(u==x) continue;
for(int e=head[u]; e!=-1; e=edge[e].next)
{
int v = edge[e].to;
if(v==x) continue;
dp[u][d] += dp[v][d-1]/du[v];
}
}
}
int main()
{
int T,u,v,m,d;
scanf("%d",&T);
while(T--)
{
init();
memset(du,0,sizeof(du));
scanf("%d%d%d",&n,&m,&d);
for(int i=0; i<m; i++)
{
scanf("%d%d",&u,&v);
addedge(u,v);
addedge(v,u);
du[u]++;
du[v]++;
}
for(int i=1; i<=n; i++)
{
dp[i][0] = 1.0/n;
ans[i]=0;
}
for(int u=1; u<=n; u++)
{
for(int j=1; j<=d; j++)
cal_dp(u,j);
for(int v=1;v<=n;v++)
ans[u] += dp[v][d];
}
for(int i=1; i<=n; i++)
{
printf("%.10lf\n",ans[i]);
}
}
}