poj 1679 The Unique MST 【次小生成树】

The 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：求最小生成树是否唯一。

什么是生成树：就是包含1~n，n个节点的一个数

什么是次小生成树，就是在图中生成树（权值）中仅仅大于最小生成树（权值）的一棵树。

分析：设T为图的一颗最小生成树，那么我们思考发现，只要我们能够找到不在T中一条边(U,V)来替换T中的（U，v）那么这个替换过的树，必定是一个生成树，又因为最小生成树中的任意两个节点的边都是相对最短的，那么我们只需要不断的换边的话（每次只换一条边），这就形成了一个由最小生成树改变得来的生成树集，叫T的邻集，那么我们不难发现，次小生成树一定是在T的邻集中，所以我们只需要枚举T的邻集找出来次小生成树，然后判断是不是与最小生成树相等即可。

那么问题来了：挖掘机技术哪家强。。咳咳 是怎么求呢？

求法：每次枚举一条在T中的边，找出在T中连接u和v的唯一路径（肯定是唯一的，，想想为啥），在路径中找出最长的那条边，用max【u】【v】储存， 然后枚举不在T中的边u， v，替换掉最长边max【u】【v】，不断枚举，

有点抽象，我说一下我的理解

图有点丑，不过能表达我的意思。MST就是最小生成树，SST就是次小生成树 u= 2， v = 3.

转换过后就是这样。

其实这道题有二种解法的，一种是求出来次小生成树，一种是标记最小生成树的边，一次删去标记的边，求最小生成树，

注:这只是我个人（小菜鸟一枚）的理解，欢迎高手来指教！！

参考：http://www.cnblogs.com/dongsheng/articles/2617186.html

 代码1（次小生成树）：

#include <cstdio>
#include <cstring>
//#include <algorithm>
const int M = 105;
const int INF = 0x3f3f3f3f;
using namespace std;

int map[M][M], n, m, low[M];
int pre[M], max[M][M];  //pre是上一个点，max就是储存最长的路径
bool con[M][M], vis[M];  con数组是标记最小生成树中的边

int Max(int a, int b){
    return a>b?a:b;
}

int prim(){
    memset(vis, 0, sizeof(vis));
    memset(max, 0, sizeof(max));
    memset(con, false, sizeof(con));
    int i, j, pos, min, sum = 0;
    pos = 1;
    for(i = 1; i <= n; i ++){
        low[i] = map[pos][i];
        pre[i] = pos;
    }
    vis[pos] = 1;
    for(i = 1; i < n; i ++){
        min = INF;
        for(j = 1; j <= n; j ++){
            if(!vis[j]&&min > low[j]){
                min = low[j]; pos = j;
            }
        }
        if(min == INF) return -1;
        sum += min;
        vis[pos] = 1;
        con[pre[pos]][pos] = con[pos][pre[pos]] = 1;
        max[pre[pos]][pos] = max[pos][pre[pos]] = min;  //先储存min
        for(j = 1; j <= n; j ++){  //这里的循环就是找出来最长边，仔细想一想哈~~用的是的dp
            max[j][pos] = Max(max[pre[pos]][pos], max[j][pos]);
        }
        for(j = 1; j <= n; j ++){
            if(!vis[j]&&map[pos][j] < low[j]){
                low[j] = map[pos][j];
                pre[j] = pos;
            }
        }
    }
    return sum;
}

int main(){
    int t;
    scanf("%d", &t);
    while(t --){
        scanf("%d%d", &n, &m);
        int i, j;
        for(i = 0; i < M; i ++)
            for(j = 0; j < M; j ++)
                map[i][j] = INF;
        int a, b, c;
        for(i = 0; i < m; i ++){
            scanf("%d%d%d", &a, &b, &c);
            map[a][b] = map[b][a] = c;
        }
        int res = prim();
        int flag = 0;
        for(i = 1; i <= n; i ++){
            for(j = 1; j <= n; j ++){
                if(con[i][j]||map[i][j] == INF) continue; //这里就是枚举最小生成树的边，
                int ans = map[i][j] - max[i][j];
                if(ans == 0){
                    flag = 1;
                    break;
                }
            }
            if(flag) break;
        }
        if(flag) printf("Not Unique!\n");
        else printf("%d\n", res);
    }
    return 0;
}

代码2：

#include <cstdio>
#include <cstring>
#include <algorithm>
const int M = 105;
using namespace std;

struct node{
    int from, to, w;
    bool flag;
}s[M*M];
int n, m, fat[M];

int f(int x){
    if(x != fat[x]) fat[x] = f(fat[x]);
    return fat[x];
}

int cmp(node a, node b){
    return a.w < b.w;
}

int kruskal(){
    int i, min = 0, cou = 0;
    for(i = 0; i < M; i ++) fat[i] = i;
    for(i = 0; i < m; i ++){
        int x = f(s[i].from); int y = f(s[i].to);
        if(x != y){
            min += s[i].w;
            fat[x] = y;
            s[i].flag = true;
            cou ++;
        }
    }
    if(cou != n-1) return -1;
    return min;
}

int kruskalsst(int v){
    int i, min = 0, cou = 0;
    for(i = 0; i < M; i ++) fat[i] = i;
    for(i = 0; i < m; i ++){
        if(v == i) continue;
        int x = f(s[i].from); int y = f(s[i].to);
        if(x != y){
            min += s[i].w;
            fat[x] = y;
            cou ++;
        }
    }
    if(cou != n-1) return -1;
    return min;
}

int main(){
    int t;
    scanf("%d", &t);
    while(t --){
        scanf("%d%d", &n, &m);
        memset(s, 0, sizeof(s));
        int i;
        for(i = 0; i < m; i ++){
            scanf("%d%d%d", &s[i].from, &s[i].to, &s[i].w);
        }
        sort(s, s+n, cmp);
        int res = kruskal();
        if(res == -1 ){
            printf("Not Unique!\n"); continue;
        }
        int flag = 0;
        for(i = 0; i < m; i ++){
            if(s[i].flag){
                int ans = kruskalsst(i);
                if(res == ans) {
                    flag = 1; break;
                }
            }
        }
        if(flag) printf("Not Unique!\n");
        else printf("%d\n", res);
    }
    return 0;
}