POJ-1639 Picnic Planning （最小度限制生成树）（模板题）

Picnic Planning

Time Limit: 5000MS Memory Limit: 10000KTotal Submissions: 8340 Accepted: 2946

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

10Alphonzo Bernardo 32Alphonzo Park 57Alphonzo Eduardo 43Bernardo Park 19Bernardo Clemenzi 82Clemenzi Park 65Clemenzi Herb 90Clemenzi Eduardo 109Park Herb 24Herb Eduardo 793

Sample Output

Total miles driven: 183

Source

East Central North America 2000

题意：

马戏团的小丑们有一个特异功能，无论一个车子有多小，他们都能钻进去，也就是说，一辆车子能够容纳无限个小丑。

现在小丑们要去一个公园野餐，他们住在不同的地方，为了节约路费（石油），要使得所有车子加起来走的路程最小，那么，小丑A

可以直接开车到公园，或者开到小丑B家，然后把车停在B家，搭B的车一起去公园。

小丑家的停车位是有限制的，但是公园的停车位是有限制的，公园最多只能停k辆车。一旦某个小丑开车到了公园，那么就必须停在公园，不能在回去载其他小丑了。

求所有小丑开车的最短总路程。

借(chao)鉴(xi)：http://blog.csdn.net/shuangde800/article/details/8038433

代码照着打的，很惭愧，就当建个模板吧。

看了

《最小生成树问题的扩展》

    http://wenku.baidu.com/view/41800d66ddccda38376bafac.html

里面定理和证明基本我都无视了，但是思路是先对整个图建立最小m度的限制生成树，m代表建图时删除v0点之后的m个连通分量。之后每通过一次删边增边（在论文里是可行交换），相对于最小m度限制生成树，构造出了最小m+1度限制生成树。由于每次删边增边，都会使整体权值减小（不增大），因此执行到k度或构造最小m+1度限制生成树而权值不变的情况下，停止即可。此时的整体权值最小，放在此题中，即总路程最短。

#include<algorithm>#include<iostream>#include<string>#include<map>#include<cstdio>#include<cstring>using namespace std;map<string, int>mp;const int VN  = 30;    // 点的数量const int EN  = VN*VN; // 边的数量const int INF = 0x7fffffff;int limit;template<typename Type>class Prim{public:    void init(int _n){        for(int i=1; i<VN-1; ++i){            w[i][i] = INF;            for(int j=i+1; j<VN; ++j)                w[i][j]=w[j][i]=INF;        }    }    void setVertexNum(int x){        n = x;    }    void insert(int u, int v, Type weight){        if(w[u][v]>weight) w[u][v] = weight; //注意可能有重复边    }    Type minDegreeST(int v0, int k){ // v0是限制度的点， k是限制的度数        memset(father, -1, sizeof(father));        memset(vis, 0, sizeof(vis));        memset(edge, 0, sizeof(edge));        vis[v0] = true;        int m = 0;  // 连通分支的个数        mst = 0;         // 所求答案        /* 步骤1: 先求出m限制树 */        for(int i=1; i<=n; ++i)if(!vis[i]){            ++m;            mst += prim(i, v0);        }        /* 步骤2: 由m限制树得到m+1限制树 */        int minAdd, a, b, tmp;        int change;  // 回路上权值最大的边，用于交换        for(int i=m+1; i<=k&&i<=n; ++i){            memset(best, -1, sizeof(best));            for(int j=1; j<=n; ++j)if(best[j]==-1 && father[j]!=v0){                Best(j, v0);            }             minAdd = INF;            for(int j=1; j<=n; ++j)if(w[v0][j]!=INF && father[j]!=v0){ //遍历所有边                a = best[j];                b = father[best[j]];                tmp = w[v0][j] - w[a][b];                if(tmp < minAdd){                    minAdd=tmp;                    change = j;                }            }            if(minAdd >= 0) break; //用于度数不大于k的限制，如果k限制，就不用break了            mst += minAdd;            a = best[change];            b = father[change];            w[a][b] = w[b][a] = INF;            father[a] = b = v0;            w[a][b] = w[b][a] = w[change][v0];            w[v0][change] = w[change][v0] = INF;        }         return mst;    }   private:    // 拉成有根树    void dfs(int cur){        for(int i=1; i<=n; ++i)if(mark[i] && edge[i][cur]){            father[i] = cur;            mark[i] = 0;            dfs(i);        }    }    // 记忆化搜索,求x到V0路径上权值最大的边    int Best(int x, int V0){        if(father[x]==V0) return -1;        if(best[x] != -1){            return best[x];        }        int tmp = Best(father[x], V0);        if(tmp!=-1 && w[tmp][father[tmp]] > w[father[x]][x])            best[x] = tmp;        else            best[x] = x;        return best[x];    }    /* 求去掉与V0相连的边之后的连通分量的最小生成树 */    Type prim(int s, int V0){        memset(mark, false, sizeof(mark));        vis[s] = mark[s] = true;        for(int i=1; i<=n; ++i){            key[i] = w[s][i]; pre[i] = s;         }        int sum=0;        for(int i=1; i<n; ++i){            int u=-1;            for(int j=1; j<=n; ++j)if(!vis[j]&&!mark[j]){                if(u==-1||key[j]<key[u]) u=j;            }            if(u==-1) break;            vis[u] = mark[u] = true;            edge[pre[u]][u] = edge[u][pre[u]] = true;            sum += w[pre[u]][u];            for(int j=1; j<=n; ++j)if(!vis[j]&&!mark[j]){                if(key[j]>w[u][j]){                    key[j] = w[u][j]; pre[j] = u;                }            }        }        int Min = INF;        int root = -1;  // 树根        for(int i=1; i<=n; ++i)if(mark[i] && w[i][V0]<Min){            Min = w[i][V0];            root = i;        }        // 拉成有根树,即把当前这个连通分量用一条到V0权值最小的边连接起来        // 并且构成一棵树,利用father数组保存父结点        mark[root] = 0;        dfs(root);        father[root] = V0;        return sum + Min;    }private:    int n;             // 结点个数    int pre[VN];       // 父结点     int father[VN];    // 生成树中的父结点    bool edge[VN][VN]; // edg[i][j] = true 表示边[i,j]已在生成树中    int best[VN];      // best[i]保存V0到i之间与V0无关联的权值最大的边    bool vis[VN];      // vis[i]表示点i是否以加入生成树    bool mark[VN];     // 用于求连通分量最小生成树的标记    Type mst;          // 保存答案     Type w[VN][VN], key[VN];};Prim<int>G;  int main(){    int n, d, a, b;    string name1,name2;    mp["Park"] = 1;    scanf("%d",&n);    G.init(n);    int cnt=1;    for(int i=0; i<n; ++i){        cin >> name1 >> name2 >> d;        a=mp[name1];  b=mp[name2];        if(!a)a=mp[name1]=++cnt;        if(!b)b=mp[name2]=++cnt;        G.insert(a,b,d);        G.insert(b,a,d);    }    G.setVertexNum(cnt);    scanf("%d",&limit);    printf("Total miles driven: %d\n", G.minDegreeST(1, limit));    return 0;}