UVA 10462 Is There A Second Way Left?（次小生成树&Prim&Kruskal）题解

思路：

Prim：

这道题目中有重边

Prim可以先加一个sec数组来保存重边的次小边，这样不会影响到最小生成树，在算次小生成树时要同时判断次小边（不需判断是否在MST中）

Kruskal：

Kruskal对重边就很友好了，不用考虑

原理是这样的：我们先找最小生成树并用used标记好哪些边是MST的边，然后我们暴力遍历每一条MST边被删去的情况，如果还能生成MST就找出这些MST最小的，这棵MST就是次小生成树

Prim代码：

#include<cmath>
#include<stack>
#include<queue>
#include<string>
#include<vector>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
const int N = 200+5;
const int INF = 0x3f3f3f3f;
using namespace std;
int n,m,p[N],Case = 1;
int mp[N][N],sec[N][N],dis[N],x[N],y[N];
int pre[N];
bool vis[N];
int Max[N][N];  //最大权值边
bool is[N][N];   //是否在MST中

void init(){
    int a,b,c;
    scanf("%d%d",&n,&m);
    memset(mp,INF,sizeof(mp));
    memset(sec,INF,sizeof(sec));
    for(int i = 1;i <= m;i++){
        scanf("%d%d%d",&a,&b,&c);
        if(c < mp[a][b]){
            sec[a][b] = sec[b][a] = min(mp[a][b],sec[a][b]);    //保存次小边
            mp[a][b] = mp[b][a] = c;
        }
        else{
            sec[a][b] = sec[b][a] = min(c,sec[a][b]);
        }
    }
    memset(vis,false,sizeof(vis));
    vis[1] = true;
    for(int i = 2;i <= n;i++){
        dis[i] = mp[i][1];
        pre[i] = 1;
    }
}
void Prim(){
    int mincost = 0;
    memset(Max,0,sizeof(Max));
    memset(is,false,sizeof(is));
    for(int i = 1;i <= n - 1;i++){
        int MIN = INF;
        int point = -1;
        for(int j = 1;j <= n;j++){
            if(!vis[j] && dis[j] < MIN){
                MIN = dis[j];
                point = j;
            }
        }
        if(point == -1){
            printf("Case #%d : No way\n",Case++);
            return;
        }
        vis[point] = true;
        mincost += MIN;
        is[point][pre[point]] = is[pre[point]][point] = true;
        for(int j = 1;j <= n;j++){
            if(vis[j] && j != point){
                Max[j][point] = Max[point][j] = max(Max[pre[point]][j],dis[point]);
            }
            if(!vis[j] && dis[j] > mp[j][point]){
                pre[j] = point;
                dis[j] = mp[j][point];
            }
        }
    }
    int seccost = INF,flag = 0;
    for(int i = 1;i <= n;i++){
        for(int j = 1;j < i;j++){
            if(mp[i][j] != INF && !is[i][j]){
                flag = 1;
                seccost = min(seccost,mincost - Max[i][j] + mp[i][j]);
            }
            if(sec[i][j] != INF){
                flag = 1;
                seccost = min(seccost,mincost - Max[i][j] + sec[i][j]);
            }
        }
    }
    if(!flag) printf("Case #%d : No second way\n",Case++);
    else printf("Case #%d : %d\n",Case++,seccost);
}
int main() {
    int T;
    scanf("%d",&T);
    while(T--){
        init();
        Prim();
    }
    return 0;
}

Kruskal代码：

#include<cmath>
#include<stack>
#include<queue>
#include<string>
#include<vector>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
const int N = 200+5;
const int INF = 0x3f3f3f3f;
using namespace std;
struct edge{
    int u,v,w;
}e[N];
bool cmp(edge a,edge b){
    return a.w < b.w;
}
int f[N],used[N];
int Case = 1;
int Find(int x){
    return f[x] == x? x : f[x] = Find(f[x]);
}
void Kruskal(){
    int n,m,a,b,c;
    scanf("%d%d",&n,&m);
    for(int i = 1;i <= m;i++){
        scanf("%d%d%d",&a,&b,&c);
        e[i].u = a;
        e[i].v = b;
        e[i].w = c;
    }
    sort(e+1,e+m+1,cmp);
    for(int i = 0;i <= n;i++) f[i] = i;
    int cnt = 0,tot = 0;
    for(int i =  1;i <= m;i++){
        int fx = Find(e[i].u);
        int fy = Find(e[i].v);
        if(fx != fy){
            f[fx] = fy;
            cnt++;
            used[tot++] = i;
        }
        if(cnt == n - 1) break;
    }
    if(cnt < n - 1){
        printf("Case #%d : No way\n",Case++);
        return;
    }
    //次小生成树
    int seccost = INF;
    for(int i = 0;i < tot;i++){
        for(int j = 0;j <= n;j++) f[j] = j;
        int num = 0,cost = 0;
        for(int j = 1;j <= m;j++){
            if(j == used[i]) continue;  //删边
            int fx = Find(e[j].u);
            int fy = Find(e[j].v);
            if(fx != fy){
                f[fx] = fy;
                num++;
                cost += e[j].w;
            }
            if(num == n - 1) break;
        }
        if(num == n-1) seccost = min(seccost,cost);
    }
    if(seccost == INF) printf("Case #%d : No second way\n",Case++);
    else printf("Case #%d : %d\n",Case++,seccost);
}
int main() {
    int T;
    scanf("%d",&T);
    while(T--){
        Kruskal();
    }
    return 0;
}