poj ces always cni can nbsp style you abi

　　　　　　　　　　　　　　　　　　　　　　　　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
一開始在網上搜索題解,照著他們的算法寫，寫完了才發現，他們有最重要的一步沒有講，幸好此時峰巨告訴了我算法的全過程，orz orz orz。

算法：
1.無視Park及其它的邊，建立最小生成樹（森林）。
2.選擇park到每個樹的最短邊，與樹相連。
3.此時，park到每棵樹還剩了一些邊，枚舉他們，每一條邊加進去都會有一個環，刪去環內的最大邊。枚舉的時候不要真實操作（或者操作後還原），而是記錄他們的值，選擇最大的，再進行刪邊操作。
4.重復第三步（k-第一步最小生成樹的數目）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)//找出環內最大邊
 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<<"次輪回"<<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<<"新增的邊 "<<j<<"--"<<p<<endl;
118             int yj=dfs(p);
119 //            cout<<"刪除邊的長度 "<<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;//還原
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<<"最終的決定 "<<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<<"初次最小生成樹   "<<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度生成樹  "<<ans<<endl;
169     printf("Total miles driven: %d\n",ans+solve(p));
170 }

POJ 1639 Picnic Planning 最小k度生成樹