标签:pac pre class tchar style color alt zoj ons
二分图+矩阵求逆
既然我们考虑b能替换哪些a,那么我们自然要得出b被哪些a表示,这里我们设一个矩阵C,那么C*A = B
为什么呢?直接A*C = B是不可行的,因为都是行向量,不能直接乘,那么我们转置一下,得出At*C=Bt,这样就很科学了,那么再转回来,A*Ct=B,于是Ct=B*A^-1那么矩阵求逆就能得出Ct,于是我们再转置回来就能得出B是由哪些A表示的,然后跑二分图匹配就行了。
二分图匹配理解的不是很清楚,大概是不能用之前已经用过较小的编号来更新自己,还是看CQzhangyu的吧。
矩阵求逆用取模就行了
#include<bits/stdc++.h> using namespace std; typedef long long ll; const int N = 305; const ll P = 999911657; int rd() { int x = 0, f = 1; char c = getchar(); 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; } ll power(ll x, ll t) { ll ret = 1; for(; t; t >>= 1, x = x * x % P) if(t & 1) ret = ret * x % P; return ret; } void up(ll &x, const ll &t) { x = ((x + t) % P + P) % P; } int n; int vis[N], Map[N][N], match[N]; struct Matrix { ll a[N][N]; Matrix() { memset(a, 0, sizeof(a)); } void set() { for(int i = 1; i <= n; ++i) a[i][i] = 1; } Matrix friend operator * (const Matrix &a, const Matrix &b) { Matrix ret; for(int i = 1; i <= n; ++i) for(int k = 1; k <= n; ++k) if(a.a[i][k]) for(int j = 1; j <= n; ++j) up(ret.a[i][j], a.a[i][k] * b.a[k][j] % P); return ret; } Matrix Inverse() { Matrix c; c.set(); for(int i = 1; i <= n; ++i) { int now = i; while(now <= n && !a[now][i]) ++now; for(int j = 1; j <= n; ++j) { swap(a[now][j], a[i][j]); swap(c.a[now][j], c.a[i][j]); } ll Inv = power(a[i][i], P - 2); for(int j = 1; j <= n; ++j) { a[i][j] = a[i][j] * Inv % P; c.a[i][j] = c.a[i][j] * Inv % P; } for(int k = 1; k <= n; ++k) if(k != i) { ll t = ((P - a[k][i] % P) % P + P) % P; for(int j = 1; j <= n; ++j) { a[k][j] = ((a[k][j] + t * a[i][j] % P) % P + P) % P; c.a[k][j] = ((c.a[k][j] + t * c.a[i][j] % P) % P + P) % P; } } } return c; } void print() { for(int i = 1; i <= n; ++i) { for(int j = 1; j <= n; ++j) printf("%d ", a[i][j]); puts(""); } } } a, b, c, d; bool dfs(int u) { for(int i = 1; i <= n; ++i) if(!vis[i] && Map[u][i]) { vis[i] = 1; if(!match[i] || dfs(match[i])) { match[i] = u; return true; } } return false; } bool dfs(int u, int lev) { for(int i = 1; i <= n; ++i) if(!vis[i] && Map[u][i]) { vis[i] = 1; if(match[i] == lev || (match[i] > lev && dfs(match[i], lev))) { match[i] = u; return true; } } return false; } int main() { // freopen("ferrous.in", "r", stdin); // freopen("ferrous.out", "w", stdout); n = rd(); for(int i = 1; i <= n; ++i) for(int j = 1; j <= n; ++j) a.a[i][j] = rd(); for(int i = 1; i <= n; ++i) for(int j = 1; j <= n; ++j) b.a[i][j] = rd(); c = b * a.Inverse(); for(int i = 1; i <= n; ++i) for(int j = 1; j <= n; ++j) if(c.a[i][j]) Map[j][i] = 1; int ans = 0; for(int i = 1; i <= n; ++i) { memset(vis, 0, sizeof(vis)); ans += dfs(i); } if(ans < n) { puts("NIE"); return 0; } for(int i = 1; i <= n; ++i) { memset(vis, 0, sizeof(vis)); dfs(i, i); } puts("TAK"); for(int i = 1; i <= n; ++i) for(int j = 1; j <= n; ++j) if(match[j] == i) printf("%d\n", j); // fclose(stdin); // fclose(stdout); return 0; }
标签:pac pre class tchar style color alt zoj ons
原文地址:http://www.cnblogs.com/19992147orz/p/7911838.html