Aeroplane chess(期望DP)
阿新 • • 發佈:2019-01-27
直接遞推就行了,如果能從x飛到y,相當於dp[x]=dp[y],特判一下就行。
AC程式碼:
#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;
}