限制某个顶点度数的最小生成树 poj1639

Picnic Planning

Time Limit: 5000MS		Memory Limit: 10000K
Total Submissions: 10642		Accepted: 3862

Description

The Contortion Brothers are a famous set of circus clowns, known worldwide for their incredible ability to cram an unlimited number of themselves into even the smallest vehicle. During the off-season, the brothers like to get together for an Annual Contortionists Meeting at a local park. However, the brothers are not only tight with regard to cramped quarters, but with money as well, so they try to find the way to get everyone to the party which minimizes the number of miles put on everyone's cars (thus saving gas, wear and tear, etc.). To this end they are willing to cram themselves into as few cars as necessary to minimize the total number of miles put on all their cars together. This often results in many brothers driving to one brother's house, leaving all but one car there and piling into the remaining one. There is a constraint at the park, however: the parking lot at the picnic site can only hold a limited number of cars, so that must be factored into the overall miserly calculation. Also, due to an entrance fee to the park, once any brother's car arrives at the park it is there to stay; he will not drop off his passengers and then leave to pick up other brothers. Now for your average circus clan, solving this problem is a challenge, so it is left to you to write a program to solve their milage minimization problem.

Input

Input will consist of one problem instance. The first line will contain a single integer n indicating the number of highway connections between brothers or between brothers and the park. The next n lines will contain one connection per line, of the form name1 name2 dist, where name1 and name2 are either the names of two brothers or the word Park and a brother's name (in either order), and dist is the integer distance between them. These roads will all be 2-way roads, and dist will always be positive.The maximum number of brothers will be 20 and the maximumlength of any name will be 10 characters.Following these n lines will be one final line containing an integer s which specifies the number of cars which can fit in the parking lot of the picnic site. You may assume that there is a path from every brother's house to the park and that a solution exists for each problem instance.

Output

Output should consist of one line of the form
Total miles driven: xxx
where xxx is the total number of miles driven by all the brothers' cars.

Sample Input

10
Alphonzo Bernardo 32
Alphonzo Park 57
Alphonzo Eduardo 43
Bernardo Park 19
Bernardo Clemenzi 82
Clemenzi Park 65
Clemenzi Herb 90
Clemenzi Eduardo 109
Park Herb 24
Herb Eduardo 79
3

Sample Output

Total miles driven: 183

以下内容均为转载（代码风格也没改，就是这么懒=_=）http://www.cnblogs.com/jackge/archive/2013/05/12/3073669.html 膜

#include<iostream>
#include<cstdio>
#include<cstring>
#include<map>
#include<climits>
#include<queue>

using namespace std;

const int N=30;

struct node{
    int v,cap;
    node(){}
    node(int _v,int _cap):v(_v),cap(_cap){}
    bool operator < (const node &a) const{
        return a.cap<cap;
    }
};

map<string,int> mp;
int g[N][N],dis[N],clo[N],pre[N],fst[N],max_side[N];
int n,m,k;

int Prim(int src,int id){
    priority_queue<node> q;
    while(!q.empty())
        q.pop();
    dis[src]=0;
    q.push(node(src,0));
    int ans=0;
    while(!q.empty()){
        node cur=q.top();
        q.pop();
        int u=cur.v;
        if(!clo[u]){
            clo[u]=id;
            ans+=dis[u];
            for(int i=1;i<n;i++)
                if(!clo[i] && g[u][i]!=0 && dis[i]>g[u][i]){ //满足松弛条件
                    pre[i]=u;
                    dis[i]=g[u][i];
                    q.push(node(i,dis[i]));
                }
        }
    }
    return ans;
}

void update(int cur,int last,int maxside){  //也是一个dfs过程,直到搜回到起点,同时完成了max_side[]更新
    max_side[cur]=maxside>g[cur][last]?maxside:g[cur][last];
    for(int i=1;i<n;i++)
        if(i!=last && g[cur][i]!=0 && (pre[cur]==i || pre[i]==cur))
            update(i,cur,max_side[cur]);
}

void Solve(){
    int i,res,cnt;
    for(i=0;i<n;i++){
        dis[i]=INT_MAX;
        clo[i]=pre[i]=fst[i]=0;
    }
    res=0,cnt=1;    //除去根节点后,图中的连通子图个数,即最小生成树个数
    for(i=1;i<n;i++)
        if(!clo[i])
            res+=Prim(i,cnt++);
    for(i=1;i<n;i++){   //找到每个生成树和 Park 最近的点使之和 Park 相连
        int id=clo[i];
        if(g[0][i]!=0 && (!fst[id] || g[0][i]<g[0][fst[id]]))
            fst[id]=i;
    }
    for(i=1;i<cnt;i++){ //把m个生成树上和根节点相连的边加入res,得到关于Park的最小m度生成树
        res+=g[0][fst[i]];
        g[0][fst[i]]=g[fst[i]][0]=0;    //之所以用邻接阵就是因为删除边很方便
        update(fst[i],0,0);
    }

    /*
    添删操作:将根节点和生成树中一个点相连,会产生一个环,将这个环上(除刚添的那条边外)权值最大
    的边删去.由于每次操作都会给总权值带来影响 d=max_side[tmp]-mat[0][tmp],我们需要得到最小生
    成树,所以我们就要求 d 尽量大
    */

    k=k-cnt+1;  //接下来重复操作,直到度数满足条件
    while(k--){
        int tmp=0;
        for(i=1;i<n;i++)    //找 d 值最大的点(就是说完成添删操作后可以使总边权减小的值最大)
            if(g[0][i]!=0 && (tmp==0 || max_side[tmp]-g[0][tmp]<max_side[i]-g[0][i]))
                tmp=i;
        if(max_side[tmp]<=g[0][tmp])    //总权值无法再减小
            break;
        res=res-max_side[tmp]+g[0][tmp];
        g[0][tmp]=g[tmp][0]=0;
        int p=0;
        for(i=tmp;pre[i]!=0;i=pre[i])
            if(p==0 || g[p][pre[p]]<g[i][pre[i]])
                p=i;
        pre[p]=0;
        update(tmp,0,0);
    }
    printf("Total miles driven: %d ",res);
}

int main(){

    //freopen("input.txt","r",stdin);
    char s1[20],s2[20];
    int cap;
    while(~scanf("%d",&m)){
        mp["Park"]=0;
        n=1;
        memset(g,0,sizeof(g));
        while(m--){
            scanf("%s %s %d",s1,s2,&cap);
            if(!mp.count(s1))
                mp[s1]=n++;
            if(!mp.count(s2))
                mp[s2]=n++;
            int u=mp[s1],v=mp[s2];
            if(!g[u][v] || g[u][v]>cap)
                g[u][v]=g[v][u]=cap;
        }
        scanf("%d",&k);
        Solve();
    }
    return 0;
}

由于一直觉得上一份有点错误 ，所以又找了一份，发现这两个思想一样啊 =_=还是觉得有点错误，先挖坑以后来填

#include <cstdio>
#include <iostream>
#include <string>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <cstring>
#include <vector>
#include <map>
using namespace std;

const int inf = 0x3f3f3f3f;
const int N = 1111;
int n,k;
map<string,int> name;
vector<string> vec;
int g[N][N],vis[N];
int li[N][N],lowcost[N],pre[N];
int kk,ans;
struct node{
 int u,v,len;
 node(){}
 node(int _u,int _v,int _del):u(_u),v(_v),len(_del){}
}del[N];
void prim(int u){
 for(int i = 1;i < n;i++){
  lowcost[i] = g[u][i];
  pre[i] = u;
 }
 vis[u] = true;
 while(true){
  int pr = -1 , mind = inf;
  for(int i = 1;i < n;i++)
   if(!vis[i] && lowcost[i] < mind){
    mind = lowcost[i];
    pr = i;
   }
  if(pr == -1) break;
  ans += mind;
  li[pre[pr]][pr] = li[pr][pre[pr]] = 1;
  if(g[0][kk] > g[0][pr]) kk = pr;
  vis[pr] = true;
  for(int i = 1;i < n;i++)
   if(!vis[i] && lowcost[i] > g[pr][i])
    lowcost[i] = g[pr][i] , pre[i] = pr;
 }
}
void dfs(int u,int fa,int del_u,int del_v){
 for(int i = 1;i < n;i++){
  if(li[u][i] && i!=fa){
   if(fa == -1 || g[del_u][del_v] < g[u][i]){
    del[i] = node(u,i,g[u][i]);
    dfs(i,u,u,i);
   }else{
    del[i] = node(del_u,del_v,g[del_u][del_v]);
    dfs(i,u,del_u,del_v);
   }
  }
 }
}
void solve(){
 memset(vis,0,sizeof(vis));
 memset(li,0,sizeof(li));
 ans = 0;
 vis[0] = true;
 for(int i = 1;i < n;i++){
  if(vis[i]) continue;
  kk = i;
  k--;
  prim(i);
  li[0][kk] = li[kk][0] = 1;
  ans += g[0][kk];
  dfs(kk,-1,-1,-1);
 }
 memset(vis,0,sizeof(vis));
 while(k--){
  int c = 0 , todel = -1;
  for(int i = 1;i < n;i++){
   if(li[0][i] || g[0][i] == inf) continue;
   if(c > g[0][i] - del[i].len){
    c = g[0][i] - del[i].len;
    todel = i;
   }
  }
  if(c == 0) break;
  ans += c;
  li[0][todel] = li[todel][0] = 1;
  li[del[todel].u][del[todel].v]=li[del[todel].v][del[todel].u] = 0;
  dfs(todel, 0 , -1,-1);
 }
 printf("Total miles driven: %d ",ans);
}
int getNum(const string &s){
 if(name.count(s) > 0)
  return name[s];
 vec.push_back(s);
 return name[s] = vec.size()-1;
}
void init(){
 memset(g,0x3f,sizeof(g));
 name.clear();
 vec.clear();
 getNum("Park");
 string su,sv;
 for(int i = 0,w;i < n;i++){
  cin >> su >> sv;
  cin >> w;
  int u = getNum(su) ,v = getNum(sv);
  if(g[u][v] > w)
   g[u][v] = g[v][u] = w;
 }
 n = vec.size();
 scanf("%d",&k);
}
int main(){
 while(scanf("%d",&n)!=EOF){
  init();
  solve();
 }
 return 0;
}