Time travel HDU - 4418 (概率DP)

对于每个点两个方向（两头只有一个方向）建一个点，然后预处理出每个点走$k(1\le k\le n)$到哪个点，列出方程式高斯消元就行了。记得前面$bfs$出那些点不可到，他们的期望没有意义。

方程式比较显然：
$E[i]=\sum_{k=1}^mp_k*(E[to(i,k)]+k)$
（$to(i,k)$表示$i$点走$k$步到的点）

#include <bits/stdc++.h>
using namespace std;
const int MAXN = 205;
const double eps = 1e-8;

struct node {
	int x, y;
	node(){}
	node(int x, int y):x(x), y(y){}
}b[MAXN];

double p[MAXN], a[MAXN][MAXN];
bool vis[MAXN];

inline bool gauss(int n) {
	for(int j = 1; j <= n; ++j) {
		if(!vis[j]) continue;
		if(!a[j][j]) {
			for(int i = j+1; i <= n; ++i)
				if(a[i][j]) {
					for(int k = j; k <= n+1; ++k)
						swap(a[i][k], a[j][k]);
					break;
				}
		}
		if(fabs(a[j][j]) < eps) return 0;
		for(int i = j+1; i <= n; ++i) if(vis[i]) {
			double v = a[i][j] / a[j][j];
			for(int k = j; k <= n+1; ++k)
				a[i][k] -= v * a[j][k];
		}
	}
	for(int i = n; i >= 1; --i) if(vis[i]) {
		for(int j = i+1; j <= n; ++j) if(vis[j])
			a[i][n+1] -= a[j][n+1] * a[i][j];
		a[i][n+1] /= a[i][i];
	}
	return 1;
}

int n, m, Y, X, D, cnt, id[MAXN][2], to[MAXN][MAXN];

queue<int>q;
inline bool bfs() {
	memset(vis, 0, sizeof vis);
	vis[id[X][D]] = 1;
	q.push(id[X][D]);
	while(!q.empty()) {
		int u = q.front(); q.pop();
		for(int i = 1; i <= m; ++i)
			if(p[i] > eps && !vis[to[u][i]])
				vis[to[u][i]] = 1, q.push(to[u][i]);
	}
	return vis[id[Y][0]] | vis[id[Y][1]];
}

int main() {
	int T;
	scanf("%d", &T);
	while(T--) {
		scanf("%d%d%d%d%d", &n, &m, &Y, &X, &D); ++X, ++Y;
		if(D == -1) D = X == 1;
		else D ^= 1;
		//0 -> leftwards
		//1 -> rightwards
		
		for(int i = 1; i <= m; ++i) scanf("%lf", &p[i]), p[i] /= 100.0;
		cnt = 0;
		id[1][0] = id[1][1] = ++cnt; b[cnt] = node(1, 1);
		id[n][0] = id[n][1] = ++cnt; b[cnt] = node(n, 0);
		for(int i = 2; i < n; ++i)
			id[i][0] = ++cnt, b[cnt] = node(i, 0),
			id[i][1] = ++cnt, b[cnt] = node(i, 1);
		
		for(int i = 1; i <= cnt; ++i) {
			node t = b[i];
			t.x += t.y ? 1 : -1;
			if(t.x == 1 || t.x == n) t.y ^= 1;
			to[i][1] = id[t.x][t.y];
		}
		for(int j = 2; j <= m; ++j)
			for(int i = 1; i <= cnt; ++i)
				to[i][j] = to[to[i][j-1]][1];
		if(!bfs()) puts("Impossible !");
		else {
			memset(a, 0, sizeof a);
			for(int i = 1; i <= cnt; ++i) {
				a[i][i] = 1;
				if(!vis[i] || i == id[Y][0] || i == id[Y][1]) continue;
				for(int j = 1; j <= m; ++j)
					a[i][to[i][j]] -= p[j],
					a[i][cnt+1] += p[j] * j;
			}
			if(!gauss(cnt)) puts("Impossible !");
			double ans = a[id[X][D]][cnt+1];
			printf("%.2f\n", ans);
		}
	}
}