POJ1639 Picnic Planning
| Time Limit: 5000MS | Memory Limit: 10000K | |
| Total Submissions: 7381 | Accepted: 2567 |
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
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题目大意：见黑书P300。
解体思路：见黑书P300。度限制最小生成树，按照黑书的意思拍代码即可。
//#pragma comment(linker, "/STACK:65536000")
#include <map>
#include <stack>
#include <queue>
#include <math.h>
#include <vector>
#include <string>
#include <fstream>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#define N 1005
#define M 200005
#define E
#define inf 0x3f3f3f3f
#define dinf 1e10
#define linf (LL)1<<60
#define LL long long
#define clr(a,b) memset(a,b,sizeof(a))
using namespace std;
int n,m,k;
int dis[N],vis[N],pd[N],pv[N],pg[N];
map<string,int>hash;
int eid,head[N],nxt[M],ed[M],val[M],flag[M];
struct Node//双重用处
{
int ed,v,id;
bool operator<(const Node & a)const
{
return v>a.v;
}
Node(int a,int b):ed(a),v(b){}
Node(int a,int b,int c):ed(a),v(b),id(c){}
};
priority_queue<Node>que;//prim优先队列
vector<Node>gra[N];//构建的最小生成树森林
void addedge(int s,int e,int v)
{
ed[eid]=e;val[eid]=v;flag[eid]=0;
nxt[eid]=head[s];head[s]=eid++;
}
//在每加一条V0带Vi的一条边时，对其他点更新
void dfs(int p,int v,int u)
{
vis[p]=1;
dis[p]=v;pd[p]=u;
int len=gra[p].size();
for(int i=0;i<len;i++)
{
int e=gra[p][i].ed,vv=gra[p][i].v,id=gra[p][i].id;
if(flag[id]||flag[id^1]||vis[e])continue;
if(vv>v)
dfs(e,vv,id);
else
dfs(e,v,u);
}
}
int main()
{
//freopen("/home/axorb/in","r",stdin);
while(scanf("%d",&m)==1)
{
hash.clear();
hash["Park"]=0;n=0;
eid=0;head[n++]=-1;
for(int i=0;i<m;i++)
{
string s1,s2;int v;
cin>>s1>>s2>>v;
if(!hash.count(s1))hash[s1]=n,head[n++]=-1;
if(!hash.count(s2))hash[s2]=n,head[n++]=-1;
int id1=hash[s1],id2=hash[s2];
addedge(id1,id2,v);
addedge(id2,id1,v);
}
cin>>k;
for(int i=0;i<n;i++)
{
dis[i]=inf;
vis[i]=0;
gra[i].clear();
}
int ans=0;
for(int i=1;i<n;i++)//构建最小生成树森林
if(dis[i]==inf)
{
int minn=inf,pos=-1,gk=-1;
while(!que.empty())que.pop();
dis[i]=0;que.push(Node(i,0));
while(!que.empty())
{
int s=que.top().ed;
que.pop();
if(vis[s])continue;
vis[s]=1;
ans+=dis[s];
if(s!=i)
{
gra[pd[s]].push_back(Node(s,pv[s],pg[s]));
gra[s].push_back(Node(pd[s],pv[s],pg[s]));
}
for(int j=head[s];~j;j=nxt[j])
{
int e=ed[j],v=val[j];
if(e==0)
{
if(minn>v)
{
minn=v;
pos=s;
gk=j;
}
continue;
}
if(!vis[e]&&dis[e]>v)
{
dis[e]=v;pd[e]=s;
pv[e]=v;pg[e]=j;
que.push(Node(e,v));
}
}
}
//这棵最小生成树选择一条与V0相关联的最小边，并更新其他点
ans+=minn;k--;flag[gk]=flag[gk^1]=1;
for(int i=0;i<n;i++)vis[i]=0;
dfs(pos,-1,-1);
}
//这里，如果k<0，则说明不存在度限制最小生成树
while(k>0)//进行最小差值运算
{
k--;
int minn=inf,pos,gk;
for(int i=head[0];~i;i=nxt[i])
{
int e=ed[i],v=val[i];
if(flag[i]||flag[i^1]||dis[e]==-1)continue;
int k=v-dis[e];
if(k<minn)
{
minn=k;
pos=e;
gk=i;
}
}
if(minn>=0)break;
ans+=minn;
flag[gk]=flag[gk^1]=1;
int g=pd[pos];
flag[g]=flag[g^1]=1;
dfs(pos,-1,-1);
}
printf("Total miles driven: %d\n",ans);
}
return 0;
}