BZOJ 2337 XOR和路径(概率DP)

求点1到点n经过的路径权值异或和的期望。

考虑按位计算，对于每一位来说，令dp[i]表示从i到n的异或和期望值。

那么dp[i]=sum(dp[j]+1-dp[k]).如果w(i,j)这一位为0，如果w(i,k)这一位为1.边界为dp[n][n]=0.

那么求解每个方程组就得到了每一位的贡献。另外注意自环的出理就ok了。

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi acos(-1.0)
# define eps 1e-9
# define MOD 1000000000
# define INF 1000000000
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<1,l,mid
# define rch p<<1|1,mid+1,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
int Scan() {
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
void Out(int a) {
    if(a<0) {putchar('-'); a=-a;}
    if(a>=10) Out(a/10);
    putchar(a%10+'0');
}
const int N=105;
//Code begin...

struct Edge{int p, next, w;}edge[20005];
int head[N], cnt=1, dee[N];
double a[N][N], x[N];
int equ, var;

void add_edge(int u, int v, int w){
    edge[cnt].p=v; edge[cnt].next=head[u]; edge[cnt].w=w; head[u]=cnt++;
}
int Guass(){
    int i, j, k, col, max_r;
    for (k=0, col=0; k<equ&&col<var; ++k, ++col){
        max_r=k;
        FO(i,k+1,equ) if (fabs(a[i][col])>fabs(a[max_r][col])) max_r=i;
        if (fabs(a[max_r][col])<eps) return 0;
        if (k!=max_r) {FO(j,col,var) swap(a[k][j],a[max_r][j]); swap(x[k],x[max_r]);}
        x[k]/=a[k][col];
        FO(j,col+1,var) a[k][j]/=a[k][col];
        a[k][col]=1;
        FO(i,0,equ) if (i!=k) {
            x[i]-=x[k]*a[i][col];
            FO(j,col+1,var) a[i][j]-=a[k][j]*a[i][col];
            a[i][col]=0;
        }
    }
    return 1;
}
int main ()
{
    int n, m, u, v, w;
    double ans=0;
    scanf("%d%d",&n,&m);
    equ=var=n;
    while (m--) {
        scanf("%d%d%d",&u,&v,&w);
        --u; --v;
        if (u!=v) add_edge(u,v,w), add_edge(v,u,w), ++dee[v];
        else add_edge(u,v,w);
        ++dee[u];
    }
    FO(i,0,32) {
        mem(a,0); mem(x,0);
        FO(j,0,n-1) {
            a[j][j]=dee[j];
            for (int k=head[j]; k; k=edge[k].next) {
                v=edge[k].p; w=edge[k].w;
                if (w&(1<<i)) {a[j][v]+=1; x[j]+=1;}
                else a[j][v]-=1;
            }
        }
        a[n-1][n-1]=1;
        Guass();
        ans+=(x[0]*(1<<i));
    }
    printf("%.3f
",ans);
    return 0;
}