Problem Description

In the unimaginable popular DotA game, the hero Lich has a wonderful skill: Frost Chain, release a jumping breath of frost that jumps N times against enemy units.

Today iSea play the role of Lich, at first he randomly chooses an enemy hero to release the skill, then the frost jumps for N times. Each time, it make a damage of one HP unit on this hero (including the first time), then bounces to another hero (can’t be himself)
if their distance is no more than D and this hero is alive of course, also randomly. Here random means equal probability.

Now we know there are always only five enemy heroes, and also their coordinates and HP value. iSea wonders the death probability of each hero. One hero is dead if its HP is equal to or less than zero.

Input

There are several test cases in the input.

Each test case begin with two integers N and D (1 <= N <= 25, 1 <= D <= 10000).

The following line contains ten integers, indicating the coordinates of the five opponents, and -10000 ≤ x, y ≤ 10000.

Then five integers follows, indicating the HP (1 <= HP <= 5) of five opponents.

The input terminates by end of file marker.

Output

For each test case, output five floating numbers, indicating the death probability of each hero, as the given order, rounded to three fractional digits, and separated by a single blank.

Sample Input

3 100
0 1 0 2 0 3 0 4 0 5
1 1 1 1 1
3 1
0 1 0 2 0 3 0 4 0 5
1 1 1 1 1

Sample Output

0.800 0.800 0.800 0.800 0.800

0.500 0.800 0.800 0.800 0.500

题意：告诉你5个敌人的位置以及他们各自的血量，你有n次机会选择距离和你小于等于d且未被杀死的敌人进行攻击，一开始先选择一个敌人进行攻击，这次攻击不算在n次攻击里面的，然后问你最后每个敌人被杀死的概率。这题网上竟搜不到题解，写了很久才写对。。

思路：开一个8维的dp，用dp[pos][shengyu][t1][t2][t3][t4][t5][idx]表示现在在第i个人身上，还剩几次打击机会，每个人剩余的血量这个状态下继续搜下去杀死每个人所加的概率。

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
typedef long long ll;
#define inf 99999999
#define maxn 150
struct node{
    int x,y;
}a[10];

int g[10][10];
int n,m;
double dp[5][26][6][6][6][6][6][6];
double ans[5];
int cas;

void init()
{
    int i,j;
    for(i=0;i<5;i++){
        for(j=0;j<5;j++){
            if((a[i].x-a[j].x)*(a[i].x-a[j].x)+(a[i].y-a[j].y)*(a[i].y-a[j].y)<=m*m  )g[i][j]=1;
            else g[i][j]=0;
        }
    }
}

void dfs(int pos,int shengyu,int t[],double p)
{
    int i,j;
    if(dp[pos][shengyu][t[0] ][t[1] ][t[2] ][t[3] ][t[4] ][5]==cas){
        for(i=0;i<5;i++){
            ans[i]+=dp[pos][shengyu][t[0] ][t[1] ][t[2] ][t[3] ][t[4] ][i]*p;
        }
        return;
    }
    if(shengyu==0){
        for(i=0;i<5;i++){
            if(t[i]==0){
                dp[pos][shengyu][t[0] ][t[1] ][t[2] ][t[3] ][t[4] ][i]=1;
                ans[i]+=p;
            }
            else{
                dp[pos][shengyu][t[0] ][t[1] ][t[2] ][t[3] ][t[4] ][i]=0;
            }
        }
        dp[pos][shengyu][t[0] ][t[1] ][t[2] ][t[3] ][t[4] ][5]=cas;
        return;
    }
    int cnt=0;
    for(i=0;i<5;i++){
        if(pos!=i && t[i] && g[pos][i])cnt++;
    }
    if(cnt==0){
        for(i=0;i<5;i++){
            if(t[i]==0){
                dp[pos][shengyu][t[0] ][t[1] ][t[2] ][t[3] ][t[4] ][i]=1;
                ans[i]+=p;
            }
            else{
                dp[pos][shengyu][t[0] ][t[1] ][t[2] ][t[3] ][t[4] ][i]=0;
            }
        }
        dp[pos][shengyu][t[0] ][t[1] ][t[2] ][t[3] ][t[4] ][5]=cas;
        return;
    }
    double c[10];
    for(i=0;i<5;i++)c[i]=ans[i];
    for(i=0;i<5;i++){
        if(i!=pos && t[i] && g[pos][i]){
            t[i]--;
            dfs(i,shengyu-1,t,p/(cnt*1.0) );
            t[i]++;
        }
    }
    dp[pos][shengyu][t[0] ][t[1] ][t[2] ][t[3] ][t[4] ][5]=cas;
    for(i=0;i<5;i++)dp[pos][shengyu][t[0] ][t[1] ][t[2] ][t[3] ][t[4] ][i]=(ans[i]-c[i])/p;
}

int main()
{
    int i,j,k,t,h,kk,qq,q;
    int b[10];
    cas=0;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        memset(ans,0,sizeof(ans));
        cas++;
        for(i=0;i<5;i++){
            scanf("%d%d",&a[i].x,&a[i].y);
        }
        for(i=0;i<5;i++){
            scanf("%d",&b[i]);
        }
        init();
        for(i=0;i<5;i++){
            b[i]--;
            dfs(i,n,b,0.2);
            b[i]++;
        }
        printf("%.3f %.3f %.3f %.3f %.3f\n",ans[0],ans[1],ans[2],ans[3],ans[4]);
    }
    return 0;
}

/*
3 100
0 1 0 2 0 3 0 4 0 5
3 4 2 1 5

0.000 0.000 0.145 0.663 0.000

4 100
0 1 0 2 0 3 0 4 0 5
3 3 3 1 1

0.015 0.015 0.015 0.784 0.784

*/