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  • BZOJ 2707: [SDOI2012]走迷宫 [高斯消元 scc缩点]

    2707: [SDOI2012]走迷宫

    题意:求s走到t期望步数,(n le 10^4),保证(|SCC| le 100)


    求scc缩点,每个scc高斯消元,scc之间直接DP

    注意每次清空系数矩阵

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <cmath>
    using namespace std;
    typedef long long ll;
    const int N=1e4+5, M=1e6+5;
    const double eps=1e-8;
    inline int read(){
        char c=getchar();int x=0,f=1;
        while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
        while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
        return x*f;
    }
    
    int n, m, s, t, de[N], u, v;
    struct edge{int v, ne;} e[M];
    int cnt=1, h[N];
    inline void ins(int u, int v) {e[++cnt]=(edge){v, h[u]}; h[u]=cnt;}
    int dfn[N], low[N], dfc, belong[N], scc;
    struct List{
    	int a[105], n;
    	int& operator [](int x) {return a[x];}
    	inline void push(int x) {a[++n]=x;}
    }li[N];
    int st[N], top;
    void dfs(int u) { //printf("dfs %d
    ",u);
    	dfn[u] = low[u] = ++dfc;
    	st[++top] = u;
    	for(int i=h[u];i;i=e[i].ne) {
    		int v=e[i].v;
    		if(!dfn[v]) dfs(v), low[u] = min(low[u], low[v]);
    		else if(!belong[v]) low[u] = min(low[u], dfn[v]);
    	}
    	if(dfn[u] == low[u]) {
    		scc++;
    		while(true) {
    			int x=st[top--];
    			belong[x] = scc;
    			li[scc].push(x);
    			if(x == u) break;
    		}
    	}
    }
    
    double a[105][105], f[N]; int id[N];
    void gauss(int n) { 
    	//puts("
    gauss");
    	//for(int i=1; i<=n; i++)
    	//	for(int j=1; j<=n+1; j++) printf("%lf%c",a[i][j], j==n+1 ? '
    ' : ' ');
    
    	for(int i=1; i<=n; i++) {
    		int r=i;
    		for(int j=i; j<=n; j++) if(abs(a[j][i])>abs(a[r][i])) r=j;
    		if(r!=i) for(int j=1; j<=n+1; j++) swap(a[r][j], a[i][j]);
    
    		for(int k=i+1; k<=n; k++) if(abs(a[k][i]) > eps){
    			double t = a[k][i]/a[i][i];
    			for(int j=i; j<=n+1; j++) a[k][j] -= t*a[i][j];
    		}
    	}
    	for(int i=n; i>=1; i--) {
    		for(int j=n; j>i; j--) a[i][n+1] -= a[i][j]*a[j][n+1];
    		a[i][n+1] /= a[i][i];
    	}
    }
    void solve(List &q) {
    	memset(a,0,sizeof(a));
    	int n=0;
    	for(int i=1; i<=q.n; i++) id[q[i]] = ++n;// printf("%d ",q[i]); puts(" q");
    	for(int i=1; i<=q.n; i++) {
    		int u=q[i]; 
    		a[i][i]=1; a[i][n+1]=1;
    		if(u==t) {a[i][n+1]=0; continue;}
    		for(int p=h[u];p;p=e[p].ne) {
    			int v=e[p].v; 
    			if(belong[v] != belong[u]) a[i][n+1] += f[v]/de[u];
    			else a[i][id[v]] -= 1.0/de[u];
    		}
    	}
    	gauss(n);
    	for(int i=1; i<=q.n; i++) f[q[i]] = a[i][n+1];// printf("getf %d %lf
    ",q[i],f[q[i]]);
    }
    namespace SCC {
    	struct edge{int v, ne;} e[M];
    	int cnt=1, h[N];
    	inline void ins(int u, int v) {e[++cnt]=(edge){v, h[u]}; h[u]=cnt;}
    
    	int vis[N];
    	void dfs(int u) {
    		if(vis[u]) return; vis[u]=1;
    		for(int i=h[u];i;i=e[i].ne) dfs(e[i].v);
    		//printf("solveSCC %d
    ",u);
    		solve(li[u]);
    	}
    }
    void build() {
    	for(int u=1; u<=n; u++)
    		for(int i=h[u];i;i=e[i].ne)
    			if(belong[u] != belong[e[i].v]) SCC::ins(belong[u], belong[e[i].v]);
    }
    int main() {
    	freopen("in","r",stdin);
    	n=read(); m=read(); s=read(); t=read();
    	for(int i=1; i<=m; i++) {
    		u=read(), v=read();
    		de[u]++; ins(u, v);
    	}
    	dfs(s);
    	//for(int i=1; i<=n; i++) printf("scc %d  %d %d
    ",i, dfn[i], belong[i]);
    	if(!dfn[t]) {puts("INF"); return 0;}
    	for(int i=1; i<=n; i++) if(i!=t && de[i]==0 && dfn[i]) {puts("INF"); return 0;}
    
    	build();
    	SCC::dfs(belong[s]);
    	//for(int i=1; i<=n; i++) printf("f %d %lf
    ",i,f[i]);
    	printf("%.3lf", f[s]);
    }
    
    
    
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  • 原文地址:https://www.cnblogs.com/candy99/p/6666726.html
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