[poj 1679] The Unique MST---(翻译：次小生成树)

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!

题目大意:给定图，判断它的最小生成树是否唯一。如果唯一的话输出最小生成树的权值和，否则输出Not Unique!

输入:
T(几组数据)
N(节点数) M(边数)
from to w( 起点,终点,权值 )
…

分析

赤裸裸的次小生成树，当次小生成树的权值与最小生成树的权值相等时，则方案不唯一

代码

#include <cstdio>#include <cstdlib>#include <cstring>#include <vector>#include <algorithm>#define open(s) freopen(s".in","r",stdin); freopen(s".out","w",stdout);#define close fclose(stdin); fclose(stdout); using namespace std;struct Edge1{    int from;    int to;    int w;    bool f;    bool operator < (const Edge1 &b)    {        return w<b.w;    }};struct Edge2{    int to;    int next;    int w;    inline void clr()    {        to=next=w=0;    }};int cnt;bool F;int n,m;int head[105];int fa[105];int size[105];int dp[105][105];bool vis[105];Edge1 side[10005];Edge2 edge[5000];vector<int>setx;inline int read(){    int k=1;    int sum=0;    char c=getchar();    for(;'0'>c || c>'9' ;c=getchar())        if(c=='-') k=-1;    for(;'0'<=c && c<='9';c=getchar())        sum=sum*10+c-'0';    return sum*k;}inline void write(int x){    if(x<0) { putchar('-'); x*=-1; }    if(x>9) write(x/10);    putchar(x%10+'0');}inline void add(int x,int y,int z)  //加边{    ++cnt;    edge[cnt].to=y;    edge[cnt].w=z;    edge[cnt].next=head[x];    head[x]=cnt;}inline int find(int x) //并查集{    return fa[x]=(fa[x]==x?x:find(fa[x]));}inline bool join(int x,int y)//并查集{    int x1=find(x),y1=find(y);    if(x1==y1) return 0;    if(size[x1]>size[y1])    {        size[x1]+=size[y1];        fa[y1]=x1;    }else    {        size[y1]+=size[x1];        fa[x1]=y1;    }    return 1;}inline int kruskal() //最小生成树{    for(int i=1;i<=n;++i)    {        fa[i]=i;        size[i]=1;    }    sort(side+1,side+m+1);    int t=n-1;    int sum=0;    for(int i=1;i<=m && t;++i)    if(join(side[i].from,side[i].to))    {        --t;        sum+=side[i].w;        add(side[i].from,side[i].to,side[i].w);        add(side[i].to,side[i].from,side[i].w);        side[i].f=1;    }    if(t) sum=-1;    return sum;}inline int max(int x,int y) //忽略{    return x>y?x:y;}inline int min(int x,int y)//忽略{    return x<y?x:y;}inline void dfs(int now) //求任意两点的最小瓶颈路{    vis[now]=1; setx.push_back(now);    for(int i=head[now];i;i=edge[i].next)    if(!vis[edge[i].to])    {        int y=edge[i].to;        for(int j=setx.size()-1;j>=0;--j)            dp[y][setx[j]]=dp[setx[j]][y]=max(dp[setx[j]][now],edge[i].w);        dfs(y);    }}inline void solve(){    n=read(); m=read();    F=1;    for(int i=1;i<=m;++i)    {        side[i].from=read();        side[i].to=read();        side[i].w=read();        side[i].f=0;    }    memset(head,0,sizeof(head));    for(;cnt;--cnt)        edge[cnt].clr();    int ans=kruskal(); //最小生成树权值和    if(ans<0) F=0;    if(F)    {        setx.clear(); //初始化        memset(dp,0,sizeof(dp));        memset(vis,0,sizeof(vis));        dfs(1); //预处理        int mi=0x3f3f3f3f;        for(int i=1;i<=m;++i)        if(!side[i].f)        {            mi=min(mi,ans+side[i].w-dp[side[i].from][side[i].to]); //次小生成树权值和            if(mi==ans){ F=0; break; }        }    }    if(F) write(ans); else printf("Not Unique!");    putchar('\n');}int main(){    open("1679");    for(int T=read();T;--T)        solve();    close;    return 0;}