1053. Path of Equal Weight (30)

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题目传送门:https://www.patest.cn/contests/pat-a-practise/1053

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 <iostream>#include <string>#include <cstdio>#include <vector>#include <queue>#include <algorithm>#include <stack>#include <map>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<Node *>tree;    for (int i=0; i<N; ++i) {        scanf("%d",&t);        Node * p  = new Node(i,t);        tree.push_back(p);    }    for (int i=0; i<M; ++i) {        scanf("%d %d",&id,&k);        tree[id]->childNum = k;        for (int j=0; j<k; ++j)        {            scanf("%d",&t);            tree[t]->parent = tree[id];        }    }        vector<vector<int>>allPath;    for (int i=0; i<N; ++i) {        int curWeight = 0;        Node * p = tree[i];        if(p->childNum)            continue;        vector<int>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<int> a,vector<int> b){        reverse(a.begin(), a.end());        reverse(b.begin(), b.end());        int len = a.size()<b.size()?(int)a.size():(int)b.size();        bool flag = false;        for (int i=0; i<len; ++i) {            if(a[i]!=b[i]){                flag = a[i]>b[i];                break;            }        }        return flag;    });    for (int i=0; i<allPath.size(); ++i) {        for (int j=(int)allPath[i].size()-1; j>=0; --j) {            printf(j==0?"%d\n":"%d ",allPath[i][j]);        }    }    return 0;}

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