BZOJ 2306 幸福路径(DP)

题解来源：http://www.cnblogs.com/jianglangcaijin/p/3799494.html

最后必然是走了一条链，或者是一个环（一直绕），或者是一条链加一个环。设f[i][j][k]表示从点j走了i步到达节点k的最大幸福度。那么f[i][j][j]就表示在绕环。那么在这个环上一直绕下去的期望为：

那么从S走i步到j再在j开始的环上绕圈的期望为：

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi acos(-1.0)
# define eps 1e-3
# define MOD 1000000000
# define INF 1000000000
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<1,l,mid
# define rch p<<1|1,mid+1,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
int Scan() {
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
void Out(int a) {
    if(a<0) {putchar('-'); a=-a;}
    if(a>=10) Out(a/10);
    putchar(a%10+'0');
}
const int N=105;
//Code begin...

double dp[N][N][N], val[N], Pow[N], sum[N];
int E[N*10][2];

int main ()
{
    int n, m, s, u, v;
    double p, ans=0;
    scanf("%d%d",&n,&m);
    FOR(i,1,n) scanf("%lf",val+i);
    scanf("%d%lf",&s,&p);
    FOR(i,1,m) scanf("%d%d",&E[i][0],&E[i][1]);
    FOR(i,0,n) FOR(j,1,n) FOR(k,1,n) dp[i][j][k]=-1e18;
    Pow[0]=1; FOR(i,1,n+1) Pow[i]=Pow[i-1]*p;
    FOR(i,1,n) sum[i]=dp[0][i][i]=val[i];
    FOR(i,1,n) FOR(j,1,n) FOR(k,1,m) dp[i][j][E[k][1]]=max(dp[i][j][E[k][1]],dp[i-1][j][E[k][0]]+val[E[k][1]]*Pow[i]);
    FOR(j,1,n) FOR(i,1,n) sum[j]=max(sum[j],(dp[i][j][j]-val[j]*Pow[i])/(1-Pow[i]));
    FOR(i,1,n) FOR(j,1,n) if (dp[i][s][j]>=0) ans=max(ans,max(dp[i][s][j],dp[i][s][j]+sum[j]*Pow[i]-val[j]*Pow[i]));
    printf("%.1f
",ans);
    return 0;
}