POJ 1639 Picnic Planning 最小k度生成树

Posted 2020-12-26 zgqblogs

　　　　　　　　　　　　　　　　　　　　　　　　Picnic Planning

Time Limit: 5000MS		Memory Limit: 10000K
Total Submissions:11615		Accepted: 4172

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

思路
一开始在网上搜索题解,照着他们的算法写，写完了才发现，他们有最重要的一步没有讲，幸好此时峰巨告诉了我算法的全过程，orz orz orz。

算法：
1.无视Park及其它的边，建立最小生成树（森林）。
2.选择park到每个树的最短边，与树相连。
3.此时，park到每棵树还剩了一些边，枚举他们，每一条边加进去都会有一个环，删去环内的最大边。枚举的时候不要真实操作（或者操作后还原），而是记录他们的值，选择最大的，再进行删边操作。
4.重复第三步（k-第一步最小生成树的数目）次

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  1 #include<iostream>
  2 #include<vector>
  3 #include<cstring>
  4 #include<algorithm>
  5 #include<cstdio>
  6 using namespace std;
  7 const int inf = 2100000000;
  8 int mp[22][22],pp;
  9 int dis[22],disx[22],n,k,kk;
 10 int book[22],flag[22][22];
 11 char name[22][515];
 12 int e1,e2,f[22];
 13 bool vis[22];
 14 int fo[22];
 15 struct node
 16 {
 17     int pio;
 18     int wei;
 19     int ex1,ex2;
 20 }getr[22],exa;
 21 
 22 
 23 int prim(int k)
 24 {
 25     dis[k]=0;
 26     int ans=0;
 27 
 28     while(true){
 29         int t=0;
 30         for(int i=1;i<=n;i++){
 31             if(!book[i]&&dis[i]<dis[t]){
 32                 t=i;
 33             }
 34         }
 35         if(t==0){break;}
 36         ans+=dis[t];book[t]=k;flag[disx[t]][t]=flag[t][disx[t]]=true;
 37         for(int i=1;i<=n;i++){
 38             if(!book[i]&&dis[i]>mp[t][i]){
 39                 dis[i]=mp[t][i];
 40                 disx[i]=t;
 41             }
 42         }
 43 
 44     }
 45     return ans;
 46 }
 47 
 48 void init()
 49 {
 50     for(int i=1;i<=22;i++){
 51         for(int j=1;j<=22;j++){
 52             mp[i][j]=inf;
 53         }
 54     }
 55     for(int i=0;i<=22;i++){
 56         dis[i]=inf;
 57     }
 58 }
 59 
 60 void scan()
 61 {
 62     int s,x,y;
 63     scanf("%d",&n);
 64     char a[15],b[15];
 65     int t=0;
 66     init();
 67     for(int i=1;i<=n;i++){
 68         scanf("%s%s%d",a,b,&s);
 69         x=y=-1;
 70         for(int j=1;j<=t;j++){
 71             if(!strcmp(name[j],a)){x=j;}
 72             if(!strcmp(name[j],b)){y=j;}
 73         }
 74         if(x==-1){x=++t;strcpy(name[t],a);}
 75         if(y==-1){y=++t;strcpy(name[t],b);}
 76         mp[x][y]=mp[y][x]=s;
 77     }
 78 
 79 
 80     scanf("%d",&k);
 81     n=t;
 82 }
 83 
 84 int dfs(int p)//找出环内最大边
 85 {
 86     vis[p]=true;
 87     int ans=-1,op;
 88 
 89     for(int i=1;i<=n;i++){
 90         if(flag[i][p]&&i==pp&&f[p]!=i){e1=i;e2=p;return mp[i][p];}
 91         if(!vis[i]&&flag[i][p]){
 92             f[i]=p;
 93             op=dfs(i);
 94 
 95             if(op!=-1){
 96                 if(op<mp[i][p]){e1=i;e2=p;return mp[i][p];}
 97                 else return op;
 98             }
 99         }
100     }
101     return ans;
102 }
103 
104 int solve(int p)
105 {
106 
107     int tx=0,ans=0;pp=p;
108     for(int i=1;i<=kk;i++){
109         tx=0;
110         memset(getr,0,sizeof(getr));
111 //        cout<<endl;
112 //        cout<<"第"<<i<<"次轮回"<<endl;
113         for(int j=1;j<=n;j++){
114             if(flag[p][j]||mp[p][j]==inf){continue;}
115             memset(vis,0,sizeof(vis));
116             flag[p][j]=flag[j][p]=true;
117 //            cout<<"新增的边 "<<j<<"--"<<p<<endl;
118             int yj=dfs(p);
119 //            cout<<"删除边的长度 "<<yj<<endl;
120             getr[tx].pio=j;getr[tx].wei=mp[p][j]-yj;
121             getr[tx].ex1=e1;getr[tx].ex2=e2;
122             tx++;
123             flag[p][j]=flag[j][p]=false;//还原
124         }
125         exa.pio=0;exa.wei=inf;
126         for(int i=0;i<n;i++){
127             if(exa.wei>getr[i].wei){exa=getr[i];}
128         }
129 //        cout<<"最终的决定 "<<exa.ex1<<" "<<exa.ex2<<" "<<exa.pio<<" "<<exa.wei<<endl;
130         ans+=min(exa.wei,0);
131         flag[p][exa.pio]=flag[exa.pio][p]=true;
132         flag[exa.ex1][exa.ex2]=flag[exa.ex2][exa.ex1]=false;
133     }
134     return ans;
135 }
136 
137 
138 int main()
139 {
140     scan();
141     int p;
142     for(int i=1;i<=n;i++){
143         if(!strcmp(name[i],"Park")){p=i;break;}
144     }
145     int m=0,ans=0;
146     book[p]=p;
147     for(int i=1;i<=n;i++){
148         if(!book[i]){
149             m++;
150             ans+=prim(i);
151         }
152     }
153 //    cout<<"初次最小生成树   "<<ans<<endl;
154     kk=k;
155     for(int i=1;i<=m;i++){
156         int minn=inf,ss=-1;;
157         for(int j=1;j<=n;j++){
158             if(!flag[p][j]&&!vis[book[j]]&&mp[p][j]!=inf){
159                 if(mp[j][p]<minn){minn=mp[j][p];ss=j;}
160             }
161         }
162         if(minn!=inf){
163             kk--;
164             ans+=mp[p][ss];vis[book[ss]]=true;
165             flag[p][ss]=flag[ss][p]=true;
166         }
167     }
168 //    cout<<"最小m度生成树  "<<ans<<endl;
169     printf("Total miles driven: %d
",ans+solve(p));
170 }