hdu5438 Ponds [DFS, STL vector two-dimensional array]

table of Contents

Topics address
Casual working
Code and explanation
reference

Topics address

Casual working

Code and explanation

Reference to answer this question when an article code blog shorter, more interesting, using a two-dimensional array STL vector.
May be understood in conjunction with the following sample code:

#include<iostream>
#include<vector>
using namespace std;
int main()
{
    vector<int> n[100];
    int i;
    for(i=0;i<100;i++){
        n[i].clear();
    }
    n[5].push_back(1);
    n[5].push_back(3);
    printf("%d\n",n[5].size());//为2
    printf("%d\n",n[5][0]);//为1
    printf("%d\n",n[5][1]);//为3
    //printf("%d\n",n[5]);//编译不通过 
    return 0;
}

Be appreciated that one-dimensional array of 100 rows, each row of variable length, vector corresponding to variable length arrays. Therefore, when the previous one-dimensional array of n ( vector<int> noperation) here [i] (for n- vector<int> n[100];int i;) applications.

Dfs dfs this question with the basic template online difference a little big, not very easy to learn as the start dfs, do not explain in detail here.
Here is the c ++ code.

#include<iostream>
#include<vector>
#include<cstring>
#include<cstdio>
using namespace std;
int v[10010],f[10010],l[10010];//v存储每个池塘的值，f存储其是否被遍历过，l存储每个池塘连接管子数
vector<int> x[10010];//x存储每个池塘与哪个池塘相连的具体情况 
int tmp;//tmp是已经遍历的节点个数 
long long sum1,sum2;
void dfs(int t);
int main()
{
    int T,p,m,a,b;//T为样例组数，p为池塘数，m为连接组合数，a、b为每组连接中两个池塘的位置（编号） 
    int i,j;
    int flag; 
    scanf("%d",&T);
    while(T--){
        scanf("%d%d",&p,&m);
        sum1=0;
        memset(v,0,sizeof(v));
        memset(f,0,sizeof(f));
        memset(l,0,sizeof(l));//初始化
        for(i=0;i<10010;i++){
            x[i].clear();//初始化 
        }
        for(i=1;i<=p;i++){//从1开始到p结束，不然会错，因为a和b存储位置是从1开始的 
            scanf("%d",&v[i]); 
        }   
        for(i=0;i<m;i++){
            scanf("%d%d",&a,&b);
            x[a].push_back(b);//a与b相连 
            x[b].push_back(a);//b与a相连 
            l[a]++;//与a相连的池塘（管子）数加1 
            l[b]++;//与b相连的池塘（管子）数加1 
        }
        flag=1;
        while(flag==1){
            flag=0;//如果所有剩下的池塘连接管子数都大于等于2，则flag就会保持为0 
            for(i=1;i<=p;i++){//i从1开始到p 
                if(l[i]==0||l[i]==1){
                    //连接管子数小于2
                     flag=1;//只要这样的池塘还存在，就让flag为1，重新执行循环 
                     f[i]=1;//表示已经遍历过
                     for(j=0;j<x[i].size();j++){//x[i].size表示位置为i的这个池塘连接的管子数 
                        l[x[i][j]]--;//让所有与位置为i的池塘相连的池塘的连接管子数减1，即删除了位置为i的池塘与其他池塘的连接关系 
                     }
                     l[i]=-1;//表示连接管子数小于2 
                }
            }
        }
        //这样处理完后就只剩下连接管子数大于等于2的池塘，接下来用dfs判断每个连接组合是否包含奇数个池塘 
        for(i=1;i<=p;i++){//i从1开始到p 
            if(f[i]==0){
                //说明是还没有遍历过的
                sum2=0;
                tmp=0;
                dfs(i);//经过这个操作，tmp变成这个组合中池塘的数量，sum2变成这个组合中所有池塘值的和，并且这个组合中所有池塘都被遍历过了 
                if(tmp%2==1){//如果包含奇数个池塘 
                    sum1+=sum2;
                }    
            } 
        }
        printf("%lld\n",sum1);
    } 
    return 0;
}
void dfs(int t){
    int i;
    tmp++;
    f[t]=1;//表示已经遍历过，这样就不会对同一个组合中的每个池塘多次计算了 
    sum2+=v[t];//加上这个池塘的值
    for(i=0;i<x[t].size();i++){
        if(f[x[t][i]]==0){//对于与这个池塘相连的所有未被遍历过的池塘 
            dfs(x[t][i]);
        }
    }
    return; 
}

When a solution of this problem and start reading the wrong meaning of the questions, add up to the final requirement is a combination of the number of ponds it is odd, and I understand became a portfolio containing an odd number of values pond.

reference

HDU 5438.Ponds【2015 ACM/ICPC Asia Regional Changchun Online】【DFS】9月13