POJ 1679 The Unique MST (Kruskal 判最小生成树是否唯一)

标签：poj   kruskal   

Description

Input

Output

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!

Source

题目链接：http://poj.org/problem?id=1679

题目大意：判断最小生成树是否唯一

题目分析：对图中的每条边，扫瞄其他边，如果存在权值相同的边则标记，然后用Kruskal求最小生成树，如果该MST中未包含标记的边，则必唯一直接输出权值，否则依次去掉这些边再求MST，若两次求得的权值相同，则说明不唯一

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int const MAXN = 105;
int const MAXM = 6000;
int m, n;
int fa[MAXN], rank[MAXN];
bool fir, flag;
struct Edge
{
    int u, v, w;
    bool equal, used, del;
}e[MAXM];

bool cmp(Edge a, Edge b)  
{  
    return a.w < b.w;  
}  

void UFset()  
{  
    for(int i = 0; i <= n; i++)  
        fa[i] = i;  
    memset(rank, 0, sizeof(rank));  
}  
  
int Find(int x)  
{  
    return x == fa[x] ? x : fa[x] = Find(fa[x]);  
}  
  
void Union(int a, int b)  
{  
    int r1 = Find(a);  
    int r2 = Find(b);  
    if(r1 == r2)  
        return;  
    if(rank[r1] > rank[r2])  
        fa[r2] = r1;  
    else  
    {  
        fa[r1] = r2;  
        if(rank[r1] == rank[r2])  
            rank[r1]++;  
    }  
}  

int Kruskal()  
{  
    int u, v;  
    int num = 0, sum = 0;  
    UFset();  
    for(int i = 0; i < m; i++)  
    {  
        if(e[i].del)
            continue;
        u = e[i].u;  
        v = e[i].v;  
        if(Find(u) != Find(v))  
        {  
            sum += e[i].w;  
            Union(u, v);  
            num++;  
            if(fir)
                e[i].used = true;
        }  
        if(num == n - 1)  
            break;  
    }  
    return sum;
} 

int main()
{
    int T, u, v, w;
    scanf("%d", &T);
    while(T--)
    {
        flag = false;
        scanf("%d %d", &n, &m);
        for(int i = 0; i < m; i++)
        {
            scanf("%d %d %d", &e[i].u, &e[i].v, &e[i].w);
            e[i].equal = e[i].used = e[i].del = false;
        }
        sort(e, e + m, cmp);
        for(int i = 0; i < m; i++)
        {
            for(int j = 0; j < m; j++)
            {
                if(i == j)
                    continue;
                if(e[i].w == e[j].w)
                    e[i].equal = true;
            }
        }
        fir = true;
        int w1 = Kruskal(), w2;
        fir = false;
        for(int i = 0; i < m; i++)
        {
            if(e[i].used && e[i].equal)
            {
                e[i].del = true;    //删去这条边
                w2 = Kruskal();
                if(w1 == w2)
                {
                    printf("Not Unique!\n");
                    flag = true;
                    break;
                }
                e[i].del = false;
            }
        }
        if(!flag)
            printf("%d\n", w1);
    }
}

POJ 1679 The Unique MST (Kruskal 判最小生成树是否唯一)

标签：poj   kruskal   

原文地址：http://blog.csdn.net/tc_to_top/article/details/43671975