BZOJ2707 [SDOI2012]走迷宮 【機率dp + tarjan + 高斯消元】

題目

Morenan被困在了一個迷宮裏。迷宮能夠視爲N個點M條邊的有向圖,其中Morenan處於起點S,迷宮的終點設爲T。惋惜的是,Morenan很是的腦小,他只會從一個點出發隨機沿着一條從該點出發的有向邊,到達另外一個點。這樣,Morenan走的步數可能很長,也多是無限,更可能到不了終點。若到不了終點,則步數視爲無窮大。但你必須千方百計求出Morenan所走步數的指望值。ios

輸入格式

第1行4個整數,N,M,S,T
第[2, M+1]行每行兩個整數o1, o2,表示有一條從o1到o2的邊。c++

輸出格式

一個浮點數,保留小數點3位,爲步數的指望值。若指望值爲無窮大,則輸出"INF"。測試

輸入樣例

9 12 1 9
1 2
2 3
3 1
3 4
3 7
4 5
5 6
6 4
6 7
7 8
8 9
9 7ui

輸出樣例

9.500spa

提示

測試點
N M
[1, 6] <=10 <=100code

[7, 12] <=200 <=10000ci

[13, 20] <=10000 <=1000000
保證強連通份量的大小不超過100
另外,均勻分佈着40%的數據,圖中沒有環,也沒有自環get

題解

此題和遊走那題有殊途同歸之妙
咱們設\(f[i]\)表示從\(i\)點出發到達終點的指望步數
就有\(f[i] = \frac{\sum (f[to] + 1)}{outde[i]}\)string

倘若這是一個DAG圖,反向dp就能夠了
可是若是是通常有向圖,咱們進行縮點,先計算後面的強聯通份量
同一個強聯通份量之間,其式子的to要麼是以後已經計算好了的點,要麼就是強聯通份量內部的點,能夠用高斯消元解出it

斷定時,只須要看S出發到達的全部點可否都到達T就能夠了

寫起來真要命
T出發的邊沒有任何意義,不要添加,不然可能會致使方程組無解

我醜陋雜亂的代碼

#include<iostream>
#include<cstdio>
#include<cmath>
#include<vector>
#include<cstring>
#include<algorithm>
#define eps 1e-8
#define LL long long int
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define BUG(s,n) for (int i = 1; i <= (n); i++) cout<<s[i]<<' '; puts("");
using namespace std;
const int maxn = 10005,maxm = 1000005,INF = 1000000000;
inline int read(){
    int out = 0,flag = 1; char c = getchar();
    while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
    while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
    return out * flag;
}
int h[maxn],ne = 2;
int h2[maxn],ne2 = 2;
int h3[maxn],ne3 = 2;
int n,m,S,T;
int inde[maxn];
double de[maxn];
struct EDGE{int to,nxt;}ed[maxm],ed2[maxm],ed3[maxm];
inline void build(int u,int v){
    if (u == T) return;
    ed[ne] = (EDGE){v,h[u]}; h[u] = ne++;
    de[u] += 1;
    ed3[ne3] = (EDGE){u,h3[v]}; h3[v] = ne3++;
}
inline void add(int u,int v){
    ed2[ne2] = (EDGE){v,h2[u]}; h2[u] = ne++;
}
int dfn[maxn],low[maxn],st[maxn],Scc[maxn],scci,top,cnt;
vector<int> scc[maxn];
void dfs(int u){
    dfn[u] = low[u] = ++cnt;
    st[++top] = u;
    Redge(u){
        if (!dfn[to = ed[k].to]){
            dfs(to);
            low[u] = min(low[u],low[to]);
        }else if (!Scc[to]) low[u] = min(low[u],dfn[to]);
    }
    if (low[u] == dfn[u]){
        scci++;
        do{
            Scc[st[top]] = scci;
            scc[scci].push_back(st[top]);
        }while (st[top--] != u);
    }
}
void tarjan(){
    for (int i = 1; i <= n; i++) if (!dfn[i]) dfs(i);
}
void rebuild(){
    for (int i = 1; i <= n; i++){
        int u = Scc[i];
        Redge(u) if (Scc[to = ed[k].to] != u)
            add(Scc[to],u);
    }
}
int vis[maxn];
void dfs1(int u){
    vis[u] = 1;
    Redge(u) if (!vis[to = ed[k].to]) dfs1(to);
}
void dfs2(int u){
    vis[u] += 2;
    for (int k = h3[u],to; k; k = ed3[k].nxt)
        if (vis[to = ed3[k].to] <= 1) dfs2(to);
}
bool check(){
    dfs1(S);
    dfs2(T);
    for (int i = 1; i <= n; i++) if (vis[i] == 1) return false;
    return true;
}
double f[maxn];
double A[205][205];
int id[maxn],c[maxn];
void gause(int u){
    int cnt = scc[u].size();
    for (int i = 1; i <= cnt; i++){
        c[i] = scc[u][i - 1];
        id[c[i]] = i;
    }
    for (int i = 1; i <= cnt + 1; i++)
        for (int j = 1; j <= cnt + 1; j++)
            if (i != j) A[i][j] = 0;
            else A[i][j] = 1;
    for (int i = 1; i <= cnt; i++){
        Redge(c[i]){
            to = ed[k].to;
            if (Scc[to] != u) A[i][cnt + 1] += (f[to] + 1) / de[c[i]];
            else A[i][id[to]] -= 1 / de[c[i]],A[i][cnt + 1] += 1 / de[c[i]];
        }
    }
    for (int i = 1; i <= cnt; i++){
        int j = i;
        while (j <= cnt && fabs(A[j][i]) <= eps) j++;
        if (j == cnt + 1){
            puts("INF");
            exit(0);
        }
        if (j != i) for (int k = 1; k <= cnt + 1; k++)
            swap(A[i][k],A[j][k]);
        for (int j = i + 1; j <= cnt; j++){
            if (fabs(A[j][i]) > eps){
                double t = A[j][i];
                for (int k = i; k <= cnt + 1; k++)
                    A[j][k] = A[j][k] / t * A[i][i];
                for (int k = i; k <= cnt + 1; k++)
                    A[j][k] -= A[i][k];
            }
        }
    }
    for (int i = cnt; i; i--){
        for (int j = i + 1; j <= cnt; j++)
            A[i][cnt + 1] -= A[i][j] * f[c[j]];
        if (fabs(A[i][i]) <= eps){
            puts("INF");
            exit(0);
        }
        A[i][cnt + 1] /= A[i][i];
        f[c[i]] = A[i][cnt + 1];
    }
}
void solve(){
    for (int i = 1; i <= scci; i++) gause(i);
    printf("%.3lf\n",f[S]);
}
int main(){
    n = read(); m = read(); S = read(); T = read();
    int a,b;
    while (m--){
        a = read(); b = read();
        build(a,b);
    }
    if (!check()){
        puts("INF");
        return 0;
    }
    tarjan();
    rebuild();
    solve();
    return 0;
}
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