BZOJ 2337: [HNOI2011]XOR和路径( 高斯消元 )

一位一位考虑异或结果, f(x)表示x->n异或值为1的概率, 列出式子然后高斯消元就行了

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#include<cstdio>

#include<cstring>

#include<algorithm>

#include<cmath>

 

using namespace std;

 

typedef long double ld;

#define b(i) (1 << (i))

 

const int maxn = 109;

 

ld mat[maxn][maxn];

int N, deg[maxn];

 

struct edge {

int to, w;

edge* next;

} E[20009], *pt = E, *head[maxn];

 

void AddEdge(int u, int v, int w) {

deg[pt->to = v]++; pt->w = w; pt->next = head[u]; head[u] = pt++;

}

 

void Init() {

memset(deg, 0, sizeof deg);

int m;

scanf("%d%d", &N, &m); N--;

for(int i = 0; i < m; i++) {

int u, v, w;

scanf("%d%d%d", &u, &v, &w);

u--; v--;

AddEdge(u, v, w);

if(u != v) AddEdge(v, u, w);

}

}

 

void Work() {

for(int i = 0; i < N; i++) {

int r = i;

for(int j = i; ++j < N; )

if(fabs(mat[j][i]) > fabs(mat[r][i])) r = j;

if(r != i) {

for(int j = 0; j <= N; j++)

swap(mat[i][j], mat[r][j]);

}

for(int j = i; ++j < N; ) {

ld t = mat[j][i] / mat[i][i];

for(int k = i; k <= N; k++)

mat[j][k] -= t * mat[i][k];

}

}

for(int i = N; i--; ) {

for(int j = i; ++j < N; ) 

mat[i][N] -= mat[i][j] * mat[j][N];

mat[i][N] /= mat[i][i];

}

}

 

int main() {

Init();

ld ans = 0;

for(int i = 0; i < 30; i++) {

memset(mat, 0, sizeof mat);

for(int j = 0; j < N; j++) {

for(edge* e = head[j]; e; e = e->next) if(e->to != N) {

if(e->w & b(i))

mat[j][e->to]--, mat[j][N]--;

else

mat[j][e->to]++;

} else if(e->w & b(i))

mat[j][N]--;

mat[j][j] -= deg[j];

}

Work();

ans += mat[0][N] * b(i);

}

printf("%.3lf ", (double) ans);

return 0;

}

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2337: [HNOI2011]XOR和路径

Time Limit: 10 Sec  Memory Limit: 128 MB
Submit: 699  Solved: 390
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Description

Input

Output

Sample Input

Sample Output

HINT

Source

Day2