先生成MST,然后对于MST上的每一条边,如果有其他边的长度与之相等,将其删去之后再求一次MST,如果和原来的cost相同,则不唯一
#include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <climits> #include <string> #include <iostream> #include <map> #include <cstdlib> #include <list> #include <set> #include <queue> #include <stack> using namespace std; typedef long long LL; const int maxn = 105; const int maxk = 105 * 105; struct Edge { int u,v,w; bool operator < (const Edge &x) const { return w < x.w; } }; Edge e[maxk]; bool del[maxk],equ[maxk],used[maxk]; int n,m,fa[maxn]; int findp(int x) { return x == fa[x] ? x : fa[x] = findp(fa[x]); } int Kruskal(bool flag) { int ret = 0; for(int i = 1;i <= n;i++) fa[i] = i; for(int i = 0;i < m;i++) if(!del[i]) { int pa = findp(e[i].u),pb = findp(e[i].v); if(pa == pb) continue; if(flag) used[i] = true; fa[pa] = pb; ret += e[i].w; } return ret; } int main() { int T; scanf("%d",&T); while(T--) { scanf("%d%d",&n,&m); memset(del,0,sizeof(del)); memset(equ,0,sizeof(equ)); memset(used,0,sizeof(used)); for(int i = 0;i < m;i++) { scanf("%d%d%d",&e[i].u,&e[i].v,&e[i].w); } sort(e,e + m); for(int i = 1;i < m;i++) { if(e[i].w == e[i - 1].w) { equ[i] = equ[i - 1] = true; } } int first = Kruskal(1); bool unique = true; for(int i = 0;i < m;i++) if(used[i] && equ[i]) { del[i] = true; int now = Kruskal(0); if(now == first) { unique = false; break; } } if(unique) printf("%d ",first); else puts("Not Unique!"); } return 0; }