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.
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 }