BZOJ 3036 綠豆蛙的歸宿【拓撲排序】【期望DP】
阿新 • • 發佈:2018-11-08
直接反圖+拓排然後跑一遍概率DP
#include <queue> #include <cmath> #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #define db double #define ll long long #define rep(i,x,y) for(ll i=(x);i<=(y);i++) #define red(i,x,y) for(ll i=(x);i>=(y);i--) using namespace std; const ll N=1e5+5; db f[N],edge[N<<1]; ll n,m,a[N],b[N]; ll cnt,to[N<<1],nxt[N<<1],head[N]; inline ll read() { ll x=0;char ch=getchar();bool f=0; while(ch>'9'||ch<'0'){if(ch=='-')f=1;ch=getchar();} while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();} return f?-x:x; } void ins(ll x,ll y,db z) { to[++cnt]=y;edge[cnt]=z;nxt[cnt]=head[x];head[x]=cnt; } void dp() { queue<ll>q; rep(i,1,n) if(!a[i]) q.push(i); while(q.size()) { ll x=q.front();q.pop(); for(ll i=head[x];i;i=nxt[i]) { ll y=to[i];db z=edge[i]; f[y]+=(f[x]+z)/b[y]; --a[y]; if(!a[y]) q.push(y); } } } int main() { n=read(),m=read(); rep(i,1,m) { ll x=read(),y=read(),z=read(); ins(y,x,z*1.0);a[x]++,b[x]++; } dp(); printf("%.2lf",f[1]); return 0; }