Intergalaxy Trips CodeForces

大意: 给定矩阵$p$, $p_{i,j}$表示每一秒点$i$到点$j$有一条边的概率, 每秒钟可以走一条边, 或者停留在原地, 求最优决策下从$1$到$n$的期望用时.

$f_x$为从$x$到$n$的期望用时, 每次肯定尽量选取$f$值小的后继走

假设每个点按$f$值排序后的序列为$a_1,a_2,...,x$, 有

$$f_x=1+f_1p_{x,a_1}+f_2p_{x,a_2}(1-p_{x,a_1})+...+f_xp_{x,x}prod(1-p_{x,a_i})$$

$$f_x=frac{1+sumlimits_{i}f_{i}p_{x,i}prodlimits_{j=1}^{i-1}(1-p_{x,a_j})}{1-prodlimits_{i}(1-p_{x,a_i})}$$

$dijkstra$求出从$n$到$1$的最短路即可. $a_i$维护分子, $b_i$维护分母.

#include <iostream>
#include <algorithm>
#include <cstdio>
#define REP(i,a,n) for(int i=a;i<=n;++i)
using namespace std;

const int N = 1e3+10;
int n,vis[N];
double a[N],b[N],p[N][N];

int main() {
	scanf("%d", &n);
	REP(i,1,n) REP(j,1,n) { 
		int t;
		scanf("%d", &t);
		p[i][j] = t/100.;
	}
	REP(i,1,n-1) a[i]=b[i]=1;
	REP(i,1,n) {
		double mi = 1e18;
		int id = 0;
		REP(j,1,n) if (!vis[j]&&b[j]<1) {
			double f = a[j]/(1-b[j]);
			if (mi>f) mi=f,id=j;
		}
		if (id==1) return printf("%.10lf
",mi),0;
		vis[id] = 1;
		REP(j,1,n) if (!vis[j]) {
			a[j]+=b[j]*p[j][id]*mi;
			b[j]*=1-p[j][id];
		}
	}
}