BZOJ 1016 生成树计数
现在给出了一个简单无向加权图。你不满足于求出这个图的最小生成树，而希望知道这个图中有多少个不同的
最小生成树。（如果两颗最小生成树中至少有一条边不同，则这两个最小生成树就是不同的）。由于不同的最小生
成树可能很多，所以你只需要输出方案数对31011的模
生成树的两个特点:
1.可能有多个生成树,但是每种权值边出现的次数在每个树中是相同的。
2.每一种生成树的每种权值边连接完成后形成的联通块状态相同。
解法1:保证了权值相同的边不超过10条 所以我们可以先得出每种权值边的个数，再暴力dfs枚举每种可能性得出这种权值的所有情况，答案就是所有情况数的累乘。
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #define mod 31011 using namespace std; int n, m, cnt, tot, ans = 1, sum; int fa[105]; struct edge { int x, y, v; } e[1005]; struct data { int l, r, v; } a[1005]; inline int read() { int x = 0; char ch = getchar(); while (ch < '0' || ch > '9') { ch = getchar(); } while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); } return x; } bool cmp(edge a, edge b) { return a.v < b.v; } int find(int x) { return x == fa[x] ? x : find(fa[x]); } void dfs(int x, int now, int k) { if (now == a[x].r + 1) { if (k == a[x].v) { sum++; } return; } int p = find(e[now].x), q = find(e[now].y); if (p != q) { fa[p] = q; dfs(x, now + 1, k + 1); fa[p] = p; fa[q] = q; } dfs(x, now + 1, k); } int main() { n = read(); m = read(); for (int i = 1; i <= n; i++) { fa[i] = i; } for (int i = 1; i <= m; i++) { e[i].x = read(), e[i].y = read(), e[i].v = read(); } sort(e + 1, e + m + 1, cmp); for (int i = 1; i <= m; i++) { if (e[i].v != e[i - 1].v) { a[++cnt].l = i; a[cnt - 1].r = i - 1; } int p = find(e[i].x), q = find(e[i].y); if (p != q) { fa[p] = q; a[cnt].v++; tot++; } } a[cnt].r = m; if (tot != n - 1) { printf("0"); return 0; } for (int i = 1; i <= n; i++) { fa[i] = i; } for (int i = 1; i <= cnt; i++) { sum = 0; dfs(i, a[i].l, 0); ans = (ans * sum) % mod; for (int j = a[i].l; j <= a[i].r; j++) { int p = find(e[j].x), q = find(e[j].y); if (p != q && a[i].v) { fa[p] = q; a[i].v--; } } } printf("%d", ans); return 0; }
解法2:矩阵树
#include<iostream> #include<algorithm> #include<cstring> #include<vector> #include<cstdio> #define p 31011 #define N 1003 using namespace std; int a[12][12],c[N][N],n,m,vis[N],fa[N],U[N]; vector<int> V[N]; struct data{ int x,y,c; bool operator<(const data &a)const{ return c<a.c; } }e[N]; int gauss(int n) { for (int i=1;i<=n;i++) for (int j=1;j<=n;j++) a[i][j]%=p; //for (int i=1;i<=n;i++,cout<<endl) // for (int j=1;j<=n;j++) cout<<a[i][j]<<" "; int ret=1; for (int i=1;i<=n;i++) { int num=i; for (int j=i+1;j<=n;j++) if (abs(a[j][i])) num=j; for (int j=1;j<=n;j++) swap(a[num][j],a[i][j]); if (num!=i) ret=-ret; for (int j=i+1;j<=n;j++) while (a[j][i]) { int t=a[j][i]/a[i][i]; for (int k=1;k<=n;k++) a[j][k]=(a[j][k]-t*a[i][k])%p; if (!a[j][i]) break; ret=-ret; for (int k=1;k<=n;k++) swap(a[i][k],a[j][k]); } ret=(ret*a[i][i])%p; } //cout<<ret<<endl; return (ret%p+p)%p; } int find(int x,int f[N]) { if (x==f[x]) return x; else return find(f[x],f); } int main() { scanf("%d%d",&n,&m); for (int i=1;i<=m;i++) scanf("%d%d%d",&e[i].x,&e[i].y,&e[i].c); sort(e+1,e+m+1); for (int i=1;i<=n;i++) vis[i]=0,fa[i]=i; int ans=1; int last=-1; for (int i=1;i<=m+1;i++) { if (e[i].c!=last||i==m+1) { for (int j=1;j<=n;j++) if (vis[j]) { int r1=find(j,U); V[r1].push_back(j); vis[j]=0; } for (int j=1;j<=n;j++) if (V[j].size()>1) { memset(a,0,sizeof(a)); int len=V[j].size(); for (int k=0;k<len;k++) for (int l=k+1;l<len;l++) { int x=V[j][k]; int y=V[j][l]; int t=c[x][y]; a[k+1][l+1]-=t; a[l+1][k+1]-=t; a[k+1][k+1]+=t; a[l+1][l+1]+=t; } ans=ans*gauss(len-1)%p; for (int k=0;k<len;k++) fa[V[j][k]]=j; } for (int j=1;j<=n;j++) { U[j]=fa[j]=find(j,fa); V[j].clear(); } last=e[i].c; if(i==m+1) break; } int x=e[i].x; int y=e[i].y; int r1=find(x,fa); int r2=find(y,fa); if (r1==r2) continue; U[find(r1,U)]=find(r2,U); vis[r1]=1; vis[r2]=1; c[r1][r2]++; c[r2][r1]++; } int flag=1; for (int i=2;i<=n;i++) if (find(i,U)!=find(i-1,U)) flag=0; ans=(ans*flag%p+p)%p; printf("%d\n",ans); }
