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枚举每个置换,求在每个置换下着色不变的方法数,先求出每个循环的大小,再动态规划求得使用给定的颜色时对应的方法数。
dp[i][j][k]表示处理到当前圈时R,B,G使用量为i,j,k时的方法数,背包思想。
#include<cstdio> #include<iostream> #include<cstdlib> #include<cstring> #include<string> #include<algorithm> #include<map> #include<queue> #include<vector> #include<cmath> #include<utility> using namespace std; typedef long long LL; const int N = 68, INF = 0x3F3F3F3F; #define MS(a, num) memset(a, num, sizeof(a)) #define PB(A) push_back(A) #define FOR(i, n) for(int i = 0; i < n; i++) int r, g, b, m, p, n; int dis[N][N]; bool vis[N]; LL dp[N][N][N]; LL Ext_gcd(LL a,LL b,LL &x,LL &y){//扩展欧几里得 if(b==0) { x=1, y=0; return a; } LL ret= Ext_gcd(b,a%b,y,x); y-= a/b*x; return ret; } LL Inv(LL a,int m){ ///求逆元 LL d,x,y,t= (LL)m; d= Ext_gcd(a,t,x,y); if(d==1) return (x%t+t)%t; return -1; } LL solve(){ LL ans = 0; for(int x = 0; x < m; x++){ memset(vis, 0, sizeof(vis)); vector<int> v; for(int i = 1; i <= n; i++){ if(!vis[i]){ int cnt = 0; int tp = i; while(!vis[tp]){ cnt++; vis[tp] = 1; tp = dis[x][tp]; } v.push_back(cnt); } } memset(dp, 0, sizeof(dp)); dp[0][0][0] = 1; for(int t = 0; t < v.size(); t++){ for(int i = r; i >= 0; i--){ for(int j = b; j >= 0; j--){ for(int k = g; k >= 0; k--){ if(i == 0 && j == 0 && k == 0){ continue; } dp[i][j][k] = 0; if(i >= v[t]){ dp[i][j][k] = (dp[i][j][k] + dp[i - v[t]][j][k]) % p; } if(j >= v[t]){ dp[i][j][k] = (dp[i][j][k] + dp[i][j - v[t]][k]) % p; } if(k >= v[t]){ dp[i][j][k] = (dp[i][j][k] + dp[i][j][k - v[t]]) % p; } } } } } ans = (ans + dp[r][b][g]) % p; } ans = ans * Inv(m, p) % p; return ans; } int main(){ while(~scanf("%d %d %d %d %d", &r, &b, &g, &m, &p)){ n = r + b + g; bool f = 1; for(int i = 0; i < m; i++){ int cnt = 0; for(int j = 1; j <= n; j++){ scanf("%d", &dis[i][j]); if(dis[i][j] == j){ cnt++; } } if(cnt == n){ f = 0; } } if(f){ for(int i = 1; i <= n; i++){ dis[m][i] = i; } m++; } printf("%lld\n", solve()); } return 0; }
BZOJ1004 [HNOI2008]Cards(Polya计数)
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原文地址:http://www.cnblogs.com/IMGavin/p/5863334.html