POJ--1679--The Unique MST【推断MST是否唯一】

链接：http://poj.org/problem?

id=1679

题意：告诉你有n个点，m条边，以及m条边的信息（起点、终点、权值）。推断最小生成树是否唯一

之前是用另外一种方法做的。复杂度最高可达O(n^3)，后来用次小生成树又做了一次。复杂度O(n^2+m)。

先说次小生成树的方法。

次小生成树：求出最小生成树，把用到的边做标记，此时加入额外的边进去必定形成环，删除环中第二大的边（即这个环里在生成树上的最大边），加上额外边的权值，枚举每一个额外边，取最小值，就是次小生成树的值。用同样的方法能够求第K小生成树以及推断第K小生成树是否唯一。

#include<cstring>
#include<string>
#include<fstream>
#include<iostream>
#include<iomanip>
#include<cstdio>
#include<cctype>
#include<algorithm>
#include<queue>
#include<map>
#include<set>
#include<vector>
#include<stack>
#include<ctime>
#include<cstdlib>
#include<functional>
#include<cmath>
using namespace std;
#define PI acos(-1.0)
#define MAXN 50100
#define eps 1e-7
#define INF 0x7FFFFFFF
#define LLINF 0x7FFFFFFFFFFFFFFF
#define seed 131
#define MOD 1000000007
#define ll long long
#define ull unsigned ll
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1

struct node{
    int u,v,w;
    int used;
}edge1[10010];
struct Edge{
    int u,v,w,next;
}edge[20010];
struct NODE{
    int u,maxm;
};
int head[110],vis[110],father[110];
int maxm[110][110];
int n,m,cnt;
void add_edge(int a,int b,int c){
    edge[cnt].u = a;
    edge[cnt].v = b;
    edge[cnt].w = c;
    edge[cnt].next = head[a];
    head[a] = cnt++;
}
int Find(int x){
    int t = x;
    while(t != father[t])
        t = father[t];
    int k = x;
    while(k != t){
        int temp = father[k];
        father[k] = t;
        k = temp;
    }
    return t;
}
int kruskal(){
    int i,j;
    int num = 0, sum = 0;
    for(i = 0; i <= n; i++) father[i] = i;
    for(i = 0; i < m; i++){
        int xx = Find(edge1[i].u);
        int yy = Find(edge1[i].v);
        if(xx != yy){
            add_edge(edge1[i].u, edge1[i].v, edge1[i].w);
            add_edge(edge1[i].v, edge1[i].u, edge1[i].w);
            father[xx] = yy;
            sum += edge1[i].w;
            edge1[i].used = 1;
            num++;
            if(num >= n - 1)    break;
        }
    }
    return sum;
}
void bfs(int src){
    int i,j;
    memset(vis, 0, sizeof(vis));
    queue<NODE>q;
    NODE t1,t2;
    t1.u = src;
    t1.maxm = 0;
    q.push(t1);
    vis[src] = 1;
    while(!q.empty()){
        t1 = q.front();
        q.pop();
        for(i = head[t1.u]; i != -1; i = edge[i].next){
            t2.u = edge[i].v;
            t2.maxm = edge[i].w;
            if(!vis[t2.u]){
                vis[t2.u] = 1;
                if(t1.maxm > t2.maxm)   t2.maxm = t1.maxm;
                maxm[src][t2.u] = t2.maxm;
                q.push(t2);
            }
        }
    }
}
int main(){
    int t,i,j;
    int a,b,c;
    scanf("%d", &t);
    while(t--){
        scanf("%d%d", &n, &m);
        for(i = 0; i < m; i++){
            scanf("%d%d%d", &edge1[i].u, &edge1[i].v, &edge1[i].w);
            edge1[i].used = 0;
        }
        memset(head,-1,sizeof(head));
        cnt = 0;
        int ans1 = kruskal();
        for(i = 1; i <= n; i++) bfs(i);
        int ans2 = INF;
        for(i = 0; i < m; i++){
            if(!edge1[i].used){
                int temp = ans1 - maxm[edge1[i].u][edge1[i].v] + edge1[i].w;
                if(temp < ans2) ans2 = temp;
            }
        }
        if(ans1!=ans2)  printf("%d\n", ans1);
        else    puts("Not Unique!");
    }
    return 0;
}

删边枚举法：对于每条边假设有和他相等权值的边。则做一个标记，然后进行一遍kruskal或prim找出最小生成树权值。然后对于每一个使用过而且有相等边标记的边，把它从图中删去，再进行一遍kruskal或prim，假设此时最小生成树权值和第一次一样，则说明最小生成树不唯一。否则最小生成树唯一。

#include<cstring>
#include<string>
#include<fstream>
#include<iostream>
#include<iomanip>
#include<cstdio>
#include<cctype>
#include<algorithm>
#include<queue>
#include<map>
#include<set>
#include<vector>
#include<stack>
#include<ctime>
#include<cstdlib>
#include<functional>
#include<cmath>
using namespace std;
#define PI acos(-1.0)
#define MAXN 110000
#define eps 1e-7
#define INF 0x7FFFFFFF
#define seed 131
#define ll long long
#define ull unsigned ll
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1

struct node{
    int u,v,dis;
    int used,equa,del;
}edge[MAXN];
int father[105],vis[105];
int n,m,flag,ans;
bool cmp(node x,node y){
    return x.dis<y.dis;
}
int find(int x){
    int t = x;
    while(father[t]!=t)
        t = father[t];
    int k = x;
    while(k!=t){
        int temp = father[k];
        father[k] = t;
        k = temp;
    }
    return t;
}
int kruskal(){
    int i,j=0;
    int sum = 0;
    for(i=1;i<=n;i++)   father[i] = i;
    for(i=0;i<m;i++){
        if(edge[i].del)     continue;
        int a = find(edge[i].u);
        int b = find(edge[i].v);
        if(a!=b){
            father[a] = b;
            sum += edge[i].dis;
            j++;
            if(flag)    edge[i].used = 1;
            if(j>=n-1)  break;
        }
    }
    return sum;
}
int main(){
    int t,i,j;
    scanf("%d",&t);
    while(t--){
        flag = 1;
        scanf("%d%d",&n,&m);
        for(i=0;i<m;i++){
            scanf("%d%d%d",&edge[i].u,&edge[i].v,&edge[i].dis);
            edge[i].used = edge[i].equa = edge[i].del = 0;
        }
        for(i=0;i<m;i++){
            for(j=0;j<m;j++){
                if(i==j)    continue;
                if(edge[i].dis == edge[j].dis)  edge[i].equa = 1;
            }
        }
        sort(edge,edge+m,cmp);
        int ans1 = kruskal();
        flag = 0;
        for(i=0;i<m;i++){
            if(edge[i].used&&edge[i].equa){
                edge[i].del = 1;
                int ans2 = kruskal();
                if(ans2==ans1){
                    printf("Not Unique!\n");
                    break;
                }
            }
        }
        if(i>=m)    printf("%d\n",ans1);
    }
    return 0;
}