POJ 1679The Unique MST

Description

Given a connected undirected graph, tell if its minimum spanning tree is unique. 
Definition 1 (Spanning Tree): Consider a connected, undirected graph G = (V, E). A spanning tree of G is a subgraph of G, say T = (V', E'), with the following properties: 
1. V' = V. 
2. T is connected and acyclic. 
Definition 2 (Minimum Spanning Tree): Consider an edge-weighted, connected, undirected graph G = (V, E). The minimum spanning tree T = (V, E') of G is the spanning tree that has the smallest total cost. The total cost of T means the sum of the weights on all the edges in E'. 

Input

The first line contains a single integer t (1 <= t <= 20), the number of test cases. Each case represents a graph. It begins with a line containing two integers n and m (1 <= n <= 100), the number of nodes and edges. Each of the following m lines contains a triple (xi, yi, wi), indicating that xi and yi are connected by an edge with weight = wi. For any two nodes, there is at most one edge connecting them.

Output

For each input, if the MST is unique, print the total cost of it, or otherwise print the string 'Not Unique!'.

Sample Input

2
3 3
1 2 1
2 3 2
3 1 3
4 4
1 2 2
2 3 2
3 4 2
4 1 2

Sample Output

3
Not Unique!

题意：t组测试数据，n个点，m条边，求最小生成树是否唯一。唯一则输出最小生成树的边劝和，否则输出Not Unque!。

没打过次小生成树，然后乱搞一通，似乎数据很水，莫名其妙的过掉了？！！
代码：

//Serene
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
using namespace std;
const int maxn=100+10,maxm=maxn*maxn;
int T,n,m,ans,mi[12];

int aa;char cc;
int read() {
	aa=0;cc=getchar();
	while(cc<'0'||cc>'9') cc=getchar();
	while(cc>='0'&&cc<='9') aa=aa*10+cc-'0',cc=getchar();
	return aa;
}

struct Line{
	int x,y,z;bool usd;
}li[maxm];

bool cmp(const Line& a,const Line& b) {return a.z<b.z;}

int fa[maxn][12];
int find(int x) {return fa[x][0]==x? x:fa[x][0]=find(fa[x][0]);}

int fir[maxn],nxt[2*maxn],to[2*maxn],e=0,v[2*maxn];
void add(int x,int y,int z) {
	to[++e]=y;nxt[e]=fir[x];fir[x]=e;v[e]=z;
	to[++e]=x;nxt[e]=fir[y];fir[y]=e;v[e]=z;
}

void Kr() {
	int tot=0,xx,yy; ans=0;
	for(int i=1;i<=n;++i) fa[i][0]=i;
	for(int i=1;i<=m&&tot<n;++i) {
		xx=find(li[i].x);yy=find(li[i].y);
		if(xx==yy) continue;
		fa[xx][0]=yy; ans+=li[i].z;
		li[i].usd=1; tot++; 
		add(li[i].x,li[i].y,li[i].z);
	}
	memset(fa,0,sizeof(fa));
}

int d[maxn],f[maxn][12];
void dfs(int pos,int dep) {
	d[pos]=dep;int y,z;
	for(y=fir[pos];y;y=nxt[y]) {
		if((z=to[y])==fa[pos][0]) continue;
		f[z][0]=v[y]; fa[z][0]=pos;
		dfs(z,dep+1);
	}
}

bool work(int x,int y,int z) {
	if(d[x]!=d[y]) {
		if(d[x]<d[y]) swap(x,y);
		int cha=d[x]-d[y];
		for(int i=10;i>=0&&cha;--i) if(cha>=mi[i]) {
			if(f[x][i]==z) return 1;
			cha-=mi[i];x=fa[x][i];
		}
	}
	if(x==y) return 0;
	for(int i=10;i>=0;--i) if(fa[x][i]!=fa[y][i]) {
		if(f[x][i]==z||f[y][i]==z) return 1;
		x=fa[x][i];y=fa[y][i];
	}
	if(f[x][1]==z||f[y][1]==z) return 1;
	return 0;
}

int main() {
	T=read(); mi[0]=1; bool ok;
	for(int i=1;i<=10;++i) mi[i]=mi[i-1]*2;
	while(T--) {
		n=read();m=read();ok=0;e=0;
		memset(fir,0,sizeof(fir));
		memset(f,0,sizeof(f));
		for(int i=1;i<=m;++i) {
			li[i].x=read();
			li[i].y=read();
			li[i].z=read();
			li[i].usd=0;
		}
		sort(li+1,li+m+1,cmp);
		Kr(); dfs(1,1);f[1][0]=1e9;
		for(int i=1;i<=10;++i) for(int j=1;j<=n;++j) {
			fa[j][i]=fa[fa[j][i-1]][i-1];
			f[j][i]=max(f[j][i-1],f[fa[j][i-1]][i-1]);
		}
		for(int i=1;i<=m;++i) if(!li[i].usd&&work(li[i].x,li[i].y,li[i].z)) {
			printf("Not Unique!\n");
			ok=1; break;
		}
		if(!ok) printf("%d\n",ans);
	}
	return 0;
}