需要明確的是 對于圖的每一個MST，所有的邊權排序后的結果都是一樣的。

先求出圖中的一個MST，對于圖中權值和MST中相同的一些邊，可以對MST中的邊替換形成新的MST，這種替換的條件是和原來的MST的這條邊構成的聯通性應該是一樣的。

可以用dfs找出對于原MST中每個權值的邊有多少種選擇。最后用乘法原理算一下即可。

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# include <cstdio> # include <cstring> # include <cstdlib> # include <iostream> # include <vector> # include <queue> # include <stack> # include <map> # include <set> # include <cmath> # include <algorithm> using namespace std; # define lowbit(x) ((x)&(-x)) # define pi acos(-1.0) # define eps 1e-3 # define MOD 1000000007 # define INF (LL)1<<60 # define mem(a,b) memset(a,b,sizeof(a)) # define FOR(i,a,n) for(int i=a; i<=n; ++i) # define FO(i,a,n) for(int i=a; i<n; ++i) # define bug puts("H"); # define lch p<<1,l,mid # define rch p<<1|1,mid+1,r # define mp make_pair # define pb push_back typedef pair<int,int> PII; typedef vector<int> VI; # pragma comment(linker, "/STACK:1024000000,1024000000") typedef long long LL; int Scan() {int res=0, flag=0;char ch;if((ch=getchar())=='-') flag=1;else if(ch>='0'&&ch<='9') res=ch-'0';while((ch=getchar())>='0'&&ch<='9') res=res*10+(ch-'0');return flag?-res:res; } void Out(int a) {if(a<0) {putchar('-'); a=-a;}if(a>=10) Out(a/10);putchar(a%10+'0'); } const int N=105; //Code begin...struct Edge{int u, v, w, flag;}edge[1005]; struct Node{int num, w;}minw[N]; int fa[N], n, m, fa1[N], p[15], now=0;int find(int x) {int s, temp;for (s=x; fa[s]>=0; s=fa[s]) ;while (s!=x) temp=fa[x], fa[x]=s, x=temp;return s; } void union_set(int x, int y) {int temp = fa[x]+fa[y];if (fa[x]>fa[y]) fa[x]=y, fa[y]=temp;else fa[y]=x, fa[x] = temp; } bool comp(Edge a, Edge b){return a.w<b.w;} bool Kruskal() {mem(fa,-1);sort(edge+1,edge+m+1,comp);int cnt=0;FOR(i,1,m) {int u=edge[i].u, v=edge[i].v, w=edge[i].w, t1=find(u), t2=find(v);if (t1!=t2) {union_set(t1,t2);++cnt;edge[i].flag=1;if (minw[now].w!=w) minw[++now].w=w, minw[now].num=1;else minw[now].num++;}if (cnt==n-1) break;}return cnt==n-1; } void copy(){FOR(i,1,n) fa[i]=fa1[i];} int dfs(int fisrt, int last, int ci, int num) {if (ci==num) {copy();FOR(i,1,num) {int u=find(edge[p[i]].u), v=find(edge[p[i]].v);if (u!=v) union_set(u,v);else return 0;}return 1;}if (fisrt>last) return 0;int ans=0;FOR(i,fisrt,last) p[ci+1]=i, ans+=dfs(i+1,last,ci+1,num);return ans; } int main () {scanf("%d%d",&n,&m);FOR(i,1,m) scanf("%d%d%d",&edge[i].u,&edge[i].v,&edge[i].w);if (Kruskal()==0) {puts("0"); return 0;}int i=1, ans=1;mem(fa,-1);FOR(j,1,now) {while (edge[i].w!=minw[j].w) ++i;int cnt=0;FOR(k,1,n) fa1[k]=fa[k];while (edge[i].w==minw[j].w) ++i, ++cnt;ans=ans*dfs(i-cnt,i-1,0,minw[j].num)%31011;copy();FOR(k,i-cnt,i-1) if (edge[k].flag) union_set(find(edge[k].u),find(edge[k].v));}printf("%d\n",ans);return 0; } View Code

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轉載于:https://www.cnblogs.com/lishiyao/p/6484377.html

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