poj1639,uva1537,uvalive2099,scu1622,fzu1761 Picnic Planning (最小限制生成树)

Posted 2020-10-02 xiepingfu

Picnic Planning

Time Limit: 5000MS		Memory Limit: 10000K
Total Submissions: 10742		Accepted: 3885

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

题意：

要求做一个最小生成树，限制条件：给定其中一个点限制其的度不超过 k （最小 k 度限制生成树）。

思路：

第一步，设被限制度数的节点为 v0 ，则在去除 v0 的情况下做最小生成树，可能得到若干个最小生成树（设有 m 棵树）；容易想到，这写树必须通过 v0 来连接成一颗树。

第二步，从 v0 引出 m 条边分别连向 m 棵树，则此时得到一个最小 m 度限制生成树，若给定的 k 小于 m 则说明这不是连通图，无法做生成树。

第三步，最多找出 k-m 条 v0 的边去替换树上现有的边；当然，替换必须使树变小才合法。这一步是比较麻烦的，并且若直接枚举的话时间复杂度也较高。每次使用动态规划找出一条贡献最大的边，并替换进树中。直到找齐 k-m 条边、或无法找到合法边是停止。此时得到的就是最小 k 度限制生成树了。

总结：

思路如上十分清晰，可实现起来细节太多了，比较坑的是同一道题不能在不同的OJ AC。在多次调试之后我的代码总算征服了poj、uva、uvalive、scu，但 fzu 却迟迟不能AC。在纵观其他大佬的题解后，发现我的代码已经算强的了....

此题需要注意的是：输入是两点之间可能存在多条边，需要保留最小的边。

代码：

#include<iostream>
#include<map>
#include<cstring>
#include<vector>
#include<queue>
#include<algorithm>
#include<cstdio>
#define READFILE freopen("D:\\in.txt","r",stdin);
#define INF 1e9+7
using namespace std;

class Edge
{
public:
    int u, v, w;
    Edge(int a=0, int b=0, int c=0):u(a), v(b), w(c) {}
};

map<string, int> mp;
vector<Edge> edges;
Edge dp[105];
int m, n, k, md, grap[105][105], fa[105], mst[105][105], ans=0;

bool cmp(Edge a, Edge b)
{
    return a.w<b.w;
}

void Init()
{
    memset(grap, -1, sizeof(grap));//-1不可达
    memset(mst, 0, sizeof(mst));
    mp.clear();
    edges.clear();
    n=1, md=0, ans=0, k=0;
    int u, v, w;
    mp["Park"]=1;

    string name1, name2;
    cin>>m;
    for(int i=0; i<m; ++i)
    {
        cin>>name1>>name2>>w;
        if(mp.find(name1)==mp.end())
            mp[name1]=n++;
        if(mp.find(name2)==mp.end())
            mp[name2]=n++;
        u=mp[name1], v=mp[name2];
        edges.push_back(Edge(u, v, w));
        if(grap[u][v]==-1)
            grap[u][v]=grap[v][u]=w;
        else
            grap[u][v]=grap[v][u]=min(grap[u][v], w);
    }
    cin>>k;
}

int Find(int x)
{
    if(fa[x]!=x)
        return fa[x]=Find(fa[x]);
    return x;
}

void Union(int a, int b)
{
    int x=Find(a);
    int y=Find(b);
    if(x!=y)
        fa[x]=y;
}

int Kruskal()//去除限制点生成md棵最小生成树
{
    int res=0;
    sort(edges.begin(), edges.end(), cmp);
    for(int i=0; i<=n; ++i)
        fa[i]=i;
    for(int i=0; i<edges.size(); ++i)
    {
        Edge& e=edges[i];
        if(e.u==1 || e.v==1 || Find(e.u)==Find(e.v)) continue;
        Union(e.u, e.v);
        mst[e.u][e.v]=mst[e.v][e.u]=grap[e.u][e.v];
        res+=grap[e.u][e.v];
    }
    return res;
}

int mmst()//生成最小md度限制生成树
{
    int minw[25], minv[25], res=0;
    for(int i=0; i<=n; ++i) minw[i]=INF;
    for(int i=2; i<=n; ++i)
        if(grap[1][i]!=-1)
        {
            int x=Find(i);
            if(minw[x] > grap[1][i])
            {
                minw[x]=grap[1][i];
                minv[x]=i;
            }
        }
    for(int i=1; i<=n; ++i)
        if(minw[i]!=INF)
        {
            md++;
            mst[1][minv[i]]=mst[minv[i]][1]=1;
            res+=grap[1][minv[i]];
        }
    return res;
}

void dfs(int x,int fa)
{
    for(int i=2; i<=n; i++)
        if(mst[x][i] && i!=fa)
        {
            if(dp[i].w==-1)
            {
                if(grap[x][i]<dp[x].w)
                {
                    dp[i].u=dp[x].u;
                    dp[i].v=dp[x].v;
                    dp[i].w=dp[x].w;
                }
                else
                    dp[i].u=x,dp[i].v=i,dp[i].w=grap[x][i];
            }
            dfs(i,x);
        }
}
int mkst()
{
    int res=0;
    for(int i=md+1; i<=k; i++)
    {
        for(int j=0; j<=n; ++j)
            dp[j].w=-1;
        dp[1].w=-INF;
        for(int j=2; j<=n; j++)
            if(mst[1][j])
                dp[j].w=-INF;
        dfs(1,-1);
        int t=0,minn=INF;
        for(int j=2; j<=n; j++)
            if(grap[1][j]!=-1&&grap[1][j]-dp[j].w<minn)
            {
                minn=grap[1][j]-dp[j].w;
                t=j;
            }
        if(minn>=0)
            break;
        mst[1][t]=mst[t][1]=1;
        int x=dp[t].u,y=dp[t].v;
        mst[x][y]=mst[y][x]=0;
        res+=minn;
    }
    return res;
}

int main()
{
    //READFILE
    int t;
    t=1;//有的oj多组数据此处改为cin>>t即可
    while(t--)
    {
        Init();
        int ans1=Kruskal();
        int ans2=mmst();
        int ans3=mkst();
        ans=ans1+ans2+ans3;
        cout<<"Total miles driven: "<<ans<<endl;
        if(t)cout<<endl;
    }
    return 0;
}