POJ1679（次小生成樹）

 

The Unique MST

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 36692		Accepted: 13368

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!






題解：
次小生成樹，維護一個兩點間的最小距離，最後再向上加

#include <stdio.h>
#include <algorithm>
#include <vector>
#include <cstring>
#include <iostream>
using namespace std;
#define  line cout<<"------------------"<<endl;
const int MAXN=1e4+10;
const int INF=0x3f3f3f3f;
int n,m;
struct node{
    int x,y;
    int v;
    bool vis;
}Edge[MAXN];
bool cmp(node a,node b)
{
    return a.v<b.v;
}
int pre[MAXN];
int Find(int a)
{
    if(pre[a]==a)
        return a;
    return Find(pre[a]);
}
vector<int >G[110];

int maxd[110][110];//並查集劃到一個樹上後，樹上任意兩點之間的距離

void init()
{
    for (int i = 1; i <=n; ++i) {
        G[i].clear();
        pre[i] = i;
        G[i].push_back(i);
    }

}
int main()
{
    int _;
    scanf("%d",&_);
    while(_--)
    {
        scanf("%d%d",&n,&m);
        init();
        for(int i=1;i<=m;i++)
        {
            scanf("%d%d%d",&Edge[i].x,&Edge[i].y,&Edge[i].v);
            Edge[i].vis=false;
        }
        sort(Edge+1,Edge+1+m,cmp);
        int sum=0;
        for (int i = 1; i <=m ; ++i) {
            int x=Find(Edge[i].x);
            int y=Find(Edge[i].y);
            if(x!=y)
            {
                pre[x]=y;
                sum+=Edge[i].v;
                int len1=G[x].size();
                int len2=G[y].size();
                for (int j = 0; j <len1 ; ++j) {
                    for (int k = 0; k <len2 ; ++k) {
                        maxd[G[x][j]][G[y][k]]=maxd[G[y][k]][G[x][j]]=Edge[i].v;//構建兩點間最小距離
                    }
                }
                int tem[110];
                for (int j = 0; j <len2 ; ++j) {
                    tem[j]=G[y][j];
                }
                for (int j = 0; j <len1 ; ++j) {
                    G[y].push_back(G[x][j]);
                }
                for (int j = 0; j <len2 ; ++j) {
                    G[x].push_back(tem[j]);
                }
                Edge[i].vis=true;
            }
        }
        int cis=INF;
        for (int i = 1; i <=m ; ++i) {//從不是最小生成樹上的邊，遍歷向上加。找到次小生成樹
           if(!Edge[i].vis)
               cis=min(cis,sum+Edge[i].v-maxd[Edge[i].x][Edge[i].y]);
        }
        if(cis>sum)
            printf("%d\n",sum);
        else
            printf("Not Unique!\n");
    }

    return 0;
}
//poj1679