POJ

POJ - 1251 Jungle Roads

The Head Elder of the tropical island of Lagrishan has a problem. A burst of foreign aid money was spent on extra roads between villages some years ago. But the jungle overtakes roads relentlessly, so the large road network is too expensive to maintain. The Council of Elders must choose to stop maintaining some roads. The map above on the left shows all the roads in use now and the cost in aacms per month to maintain them. Of course there needs to be some way to get between all the villages on maintained roads, even if the route is not as short as before. The Chief Elder would like to tell the Council of Elders what would be the smallest amount they could spend in aacms per month to maintain roads that would connect all the villages. The villages are labeled A through I in the maps above. The map on the right shows the roads that could be maintained most cheaply, for 216 aacms per month. Your task is to write a program that will solve such problems.

Input
The input consists of one to 100 data sets, followed by a final line containing only 0. Each data set starts with a line containing only a number n, which is the number of villages, 1 < n < 27, and the villages are labeled with the first n letters of the alphabet, capitalized. Each data set is completed with n-1 lines that start with village labels in alphabetical order. There is no line for the last village. Each line for a village starts with the village label followed by a number, k, of roads from this village to villages with labels later in the alphabet. If k is greater than 0, the line continues with data for each of the k roads. The data for each road is the village label for the other end of the road followed by the monthly maintenance cost in aacms for the road. Maintenance costs will be positive integers less than 100. All data fields in the row are separated by single blanks. The road network will always allow travel between all the villages. The network will never have more than 75 roads. No village will have more than 15 roads going to other villages (before or after in the alphabet). In the sample input below, the first data set goes with the map above.
Output
The output is one integer per line for each data set: the minimum cost in aacms per month to maintain a road system that connect all the villages. Caution: A brute force solution that examines every possible set of roads will not finish within the one minute time limit.
Sample Input
9
A 2 B 12 I 25
B 3 C 10 H 40 I 8
C 2 D 18 G 55
D 1 E 44
E 2 F 60 G 38
F 0
G 1 H 35
H 1 I 35
3
A 2 B 10 C 40
B 1 C 20
0
Sample Output
216
30

嘿嘿嘿，从今天开始写博客啦！
接下来应该会每天更新，顺便复习一些算法再补补题。

**咳咳咳，进入正题，这题是一道典型的最小生成树问题，题意大概就是讲有n个点，然后有n-1个行描述关系，每行第一个数字表示有多少条线与前面的点相连，后面的数字则为边权值。
这题可以采用prim算法或者是kruskal算法来解决，算是模板题啦。

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;int par[200005];int rak[200005];int cnt;struct edge{    int u,v,cost;//u，v表示点，cost表示权值};edge es[155];//并查集初始化void init(int n){    for(int i=0;i<=n;i++)    {        par[i]=i;        rak[i]=0;    }}int find(int x){    if(par[x]==x)        return x;    else        return par[x]=find(par[x]);}void unite(int x,int y){    x=find(x);    y=find(y);    if(x==y)        return ;    if(rak[x]<rak[y])    {        par[x]=y;    }    else    {        par[y]=x;    }    if(rak[x]==rak[y])        rak[x]++;}bool same(int x,int y){    return find(x)==find(y);}bool cmp(const edge&e1,const edge&e2){    return e1.cost<e2.cost;}int kruskal(){    sort(es,es+cnt,cmp);    //cout<<cnt;    init(155);    int res=0;    for(int i=0;i<cnt;i++)    {        edge e=es[i];        //printf("%d %d\n",e.u,e.v);        if(!same(e.u,e.v))        {            unite(e.v,e.u);            res+=e.cost;            //printf("%d\n",res);        }    }    return res;}int main(){    char c[10];    int n;    int num;    while(scanf("%d",&n)!=EOF)    {        cnt=0;        //init(105);        if(n<=0)            break;        for(int i=0;i<n-1;i++)        {            scanf("%s %d",c,&num);            int u=c[0]-'A';            for(int j=0;j<num;j++)            {                int b;                scanf("%s %d",c,&b);                int v=c[0]-'A';                es[cnt].u=u;                es[cnt].v=v;                es[cnt++].cost=b;                //printf("%d %d %d\n",u,v,b);                //kruskal(n);                //printf("%d\n",kruskal(n));            }        }        printf("%d\n",kruskal());    }}

补这道题的时候本以为很快就能AC，可是第一组测试数据一直是0，
调试了大概半个小时才发现原来我在并查集的初始化里将cnt也初始化了，导致了kruskal根本就没有运算，呜呜呜，下次一定要注意了。