【杭电oj】4405 - Aeroplane chess（期望DP）

题目链接

直接递推就行了，如果能从x飞到y，相当于dp[x]=dp[y]，特判一下就行。

AC代码：

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>
#include <ext/rope>
using namespace std;
using namespace __gnu_cxx;
typedef long long ll;
typedef pair<int, int> pii;
inline int read(){int r=0;char c=getchar();while(c<'0'||c>'9') {c=getchar();}while(c>='0'&&c<='9') {r=r*10+c-'0';c=getchar();}return r;}
inline ll readll(){ll r=0;char c=getchar();while(c<'0'||c>'9') {c=getchar();}while(c>='0'&&c<='9') {r=r*10+c-'0';c=getchar();}return r;}
const double eps = 1e-8;
const double PI = acos(-1.0);
const ll INFLL = 0x3f3f3f3f3f3f3f3f;
const int INF = 0x3f3f3f3f;
const int mod = 1e9+7;
const int MAXN = 1e5;
const int MAXM = 1e5;

double dp[MAXN];
int fly[MAXN];
int main(int argc, char const *argv[])
{
    int n, m;
    while(cin >> n >> m && n+m){
        memset(dp, 0, sizeof dp);
        memset(fly, 0, sizeof fly);
        int x, y;
        for(int i=0; i<m; ++i){
            cin >> x >> y;
            fly[x] = y;
        }
        for(int i=n-1; i>=0; --i){
            if(fly[i]){
                dp[i] = dp[fly[i]];
            } else{
                dp[i] = 1.0/6*(dp[i+1]+dp[i+2]+dp[i+3]+dp[i+4]+dp[i+5]+dp[i+6]) + 1.0;
            }
        }
        printf("%.4lf\n", dp[0]);
    }
    return 0;
}