POJ 1639:Picnic Planning(最小度限制生成树）
Picnic Planning
| Time Limit: 5000MS | Memory Limit: 10000K | |
| Total Submissions: 7356 | Accepted: 2555 |
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
Source
East Central North America 2000
题意：一共有nv个人从家里出发要到目的地的“Park”，一个人可以开车去接另一个人，一辆车可载重无限，问这n个人到Park的总路程最小，此外，还有一个条件，Park处最多能挺s辆车，即Park的度数不能超过s，求最小生成树。
大概思路：设Park为标号0，其他地点为1,2...，n-1；先求出1~n-1的最小生成树，然后在枚举0结点度数d从1~s的情况，即向已生成的最小生成树插入边，当d>=2的时候，必然产生回路，因此求出回路中最大的边，进行判断是否替换……
源代码：（0Ms）
#include<iostream>
#include<cstring>
using namespace std;
const int MAX_NV = 21;
const int INF = 0x7f7f7f7f;
typedef struct Edge
{
int sv,ev,w;
}Edge;
int nv;
char name[MAX_NV][12];
int gam[MAX_NV][MAX_NV];
Edge mstEdge[MAX_NV];
int s;
int ans;
bool isCycle;
int IndName(char ch[])
{
int ind=0;
while(ind<nv && strcmp(name[ind],ch)!=0)
ind++;
if(ind==nv)
strcpy(name[nv++],ch);
return ind;
}
int Prim()
{
int res=0;
int i,j,k;
for(i=1;i<nv-1;i++)
{
mstEdge[i].sv = 1;
mstEdge[i].ev = i+1;
mstEdge[i].w = gam[1][i+1];
}
for(k=2;k<nv;k++)
{
int minw = mstEdge[k-1].w,ind = k-1;
for(i=k;i<nv-1;i++)
if(minw > mstEdge[i].w)
{
minw = mstEdge[i].w;
ind = i;
}
res += minw;
Edge tmp = mstEdge[ind]; mstEdge[ind] = mstEdge[k-1]; mstEdge[k-1]=tmp;
j = mstEdge[k-1].ev;
for(i=k;i<nv-1;i++)
{
int v = mstEdge[i].ev,w = gam[j][v];
if(mstEdge[i].w > w)
{
mstEdge[i].w = w;
mstEdge[i].sv = j;
}
}
}
return res;
}
void MaxWeightEdgeInCycle(int mv,int sv,int ev,int& maxw,int& ind)
{
if(mv == ev)
{
isCycle = true;
return;
}
for(int i=0;i<nv-1;i++)
{
if(mstEdge[i].sv != ev && mstEdge[i].ev != ev)
continue;
if(mstEdge[i].sv == ev && mstEdge[i].ev != sv)
{
MaxWeightEdgeInCycle(mv,ev,mstEdge[i].ev,maxw,ind);
if(isCycle)
{
if(maxw<mstEdge[i].w && mstEdge[i].ev!=0)
{
maxw=mstEdge[i].w;
ind=i;
}
break;
}
}
else if(mstEdge[i].sv != sv && mstEdge[i].ev == ev)
{
MaxWeightEdgeInCycle(mv,ev,mstEdge[i].sv,maxw,ind);
if(isCycle)
{
if(maxw<mstEdge[i].w && mstEdge[i].sv!=0)
{
maxw=mstEdge[i].w;
ind=i;
}
break;
}
}
}
}
void Solve()
{
int i;
bool exist[MAX_NV];
ans = Prim();
int minw = INF+1,ev = -1;
for(i=1;i<nv;i++)
{
if(gam[0][i] < minw)
{
minw = gam[0][i];
ev = i;
}
exist[i]=false;
}
ans += minw;
exist[ev]=true;
mstEdge[0].w=minw; mstEdge[0].sv=0; mstEdge[0].ev=ev;
for(int d=2;d<=s;d++)
{
int dec = INF+1,edgeInd=-1;
ev = -1;
for(i=1;i<nv;i++)
{
if(exist[i]==true)
continue;
int maxw=0,ind=-1;
isCycle = false;
MaxWeightEdgeInCycle(0,0,i,maxw,ind);
if(dec > gam[0][i]-maxw)
{
dec=gam[0][i]-maxw;
edgeInd=ind;
ev=i;
}
}
if(dec>=0)
break;
else
{
mstEdge[edgeInd].sv=0; mstEdge[edgeInd].ev=ev; mstEdge[edgeInd].w=gam[0][ev];
ans += dec;
exist[ev]=true;
}
}
}
int main()
{
int i;
char name1[12],name2[12];
int ne;
strcpy(name[0],"Park");
memset(gam,0x7f,sizeof(gam));
while(scanf("%d",&ne)!=EOF)
{
nv = 1;
int dis,ind1,ind2;
for(i=0;i<ne;i++)
{
cin >> name1 >> name2 >> dis;
ind1=IndName(name1);
ind2=IndName(name2);
gam[ind1][ind2] = gam[ind2][ind1] = dis;
}
cin >> s;
Solve();
printf("Total miles driven: %d\n",ans);
}
return 0;
}