1053. Path of Equal Weight (30)-白红宇

1053. Path of Equal Weight (30)

阅读量：4072 次

发布时间：2019-05-25

本文共 1060 字，大约阅读时间需要 3 分钟。

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AC代码

//不难但一定要细心,要求是叶子节点到根的weight之和,叶子节点根,缺一不可.

#include 
   
    #include 
    
     #include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          using namespace std;struct Node{ int weight,id,childNum; Node * parent; Node(int _id,int _weight):weight(_weight),id(_id),parent(NULL),childNum(0){};};int main(){ int N,M,t,targetWeight,id,k; scanf("%d %d %d",&N,&M,&targetWeight); vector
           
            tree; for (int i=0; i
            
             childNum = k; for (int j=0; j
             
              parent = tree[id]; } } vector
              
               
                >allPath; for (int i=0; i
                
                 childNum) continue; vector
                 
                  curPath; while (true) { curWeight = curWeight + p->weight; curPath.push_back(p->weight); if(curWeight ==targetWeight){ if(p==tree[0]) allPath.push_back(curPath); break; } else if(curWeight>targetWeight) break; p = p->parent; if(!p) break; } } sort(allPath.begin(), allPath.end(), [](vector
                  
                    a,vector
                   
                     b){ reverse(a.begin(), a.end()); reverse(b.begin(), b.end()); int len = a.size()
                    
                     b[i]; break; } } return flag; }); for (int i=0; i
                     
                      =0; --j) { printf(j==0?"%d\n":"%d ",allPath[i][j]); } } return 0;}