[HNOI2011]XOR和路径

嘟嘟嘟


一看到异或，就想到按位处理．
当处理到第(i)位的时候，(f[u])表示节点(u)到(n)的路径，这一位为(1)的期望，那么为(0)就是(1 - f[u])，于是有

[f[u] = frac{1}{d[u]} (sum _ {v in V, w = 0} f[v] + sum _ {v in V, w = 1} 1 - f[v]) ]

因为是异或，所以如果边权这一位是0的话，应该加上(f[v])；否则加上(1 - f[v])。
然后整理一下

[d[u] * f[u] - sum _ {v in V, w = 0} f[v] + sum _ {v in V, w = 1} f[v] = sum _ {v in V, w = 1} 1 ]

于是就可以高斯消元了。
答案为(sum 2 ^ i * ans_i[1])。


需要注意的是,重边只应该加一次,对应的度数也应该只加(1)。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("") 
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define rg register
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 105;
const int maxe = 2e4 + 5;
inline ll read()
{
  ll ans = 0;
  char ch = getchar(), last = ' ';
  while(!isdigit(ch)) last = ch, ch = getchar();
  while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
  if(last == '-') ans = -ans;
  return ans;
}
inline void write(ll x)
{
  if(x < 0) x = -x, putchar('-');
  if(x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}

int n, m, Max = 0;
int du[maxn];
struct Edge
{
  int nxt, to, w;
}e[maxe];
int head[maxn], ecnt = -1;
void addEdge(int x, int y, int w)
{
  e[++ecnt] = (Edge){head[x], y, w};
  head[x] = ecnt;
}

db f[maxn][maxn], ans[maxn], Ans = 0;
void build(int x)
{
  Mem(f, 0);
  for(int i = 1; i < n; ++i)  //小于n
    {
      f[i][i] = du[i];
      for(int j = head[i]; j != -1; j = e[j].nxt)
	if((e[j].w >> x) & 1) ++f[i][e[j].to], ++f[i][n + 1];
	else --f[i][e[j].to];
    }
}
db Gauss()
{
  for(int i = 1; i <= n; ++i)
    {
      int pos = i;
      for(int j = i + 1; j <= n; ++j)
	if(fabs(f[j][i]) > fabs(f[pos][i])) pos = j;
      if(pos != i) swap(f[i], f[pos]);
      db tp = f[i][i];
      if(fabs(tp) > eps) for(int j = i; j <= n + 1; ++j) f[i][j] /= tp;
      for(int j = i + 1; j <= n; ++j)
	{
	  db tp = f[j][i];
	  for(int k = i; k <= n + 1; ++k) f[j][k] -= tp * f[i][k];
	}
    }
  for(int i = n; i; --i)
    {
      ans[i] = f[i][n + 1];
      for(int j = i - 1; j; --j) f[j][n + 1] -= f[j][i] * f[i][n + 1];
    }
  return ans[1];
}

int main()
{
  Mem(head, -1);
  n = read(); m = read();
  for(int i = 1; i <= m; ++i)
    {
      int x = read(), y = read(), w = read();
      addEdge(x, y, w); ++du[x];
      if(x ^ y) addEdge(y, x, w), ++du[y];
      Max = max(Max, w);
    }
  for(int i = 0; (1 << i) <= Max; ++i)
    build(i), Ans += Gauss() * (1 << i);
  printf("%.3lf
", Ans);
  return 0;
}