1053. Path of Equal Weight

最新推荐文章于 2022-04-13 00:26:50 发布

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最新推荐文章于 2022-04-13 00:26:50 发布

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本文介绍了一个寻找树中所有等权重路径的算法问题。通过DFS深度优先搜索遍历树的节点，寻找从根到叶子节点的所有路径，这些路径的权重之和等于给定的数值。文章提供了完整的代码实现，并解释了如何正确比较不同路径。

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1053. Path of Equal Weight (30)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

Given a non-empty tree with root R, and with weight W_i assigned to each tree node T_i. The weight of a path from R to L is defined to be the sum of the weights of all the nodes along the path from R to any leaf node L.

Now given any weighted tree, you are supposed to find all the paths with their weights equal to a given number. For example, let's consider the tree showed in Figure 1: for each node, the upper number is the node ID which is a two-digit number, and the lower number is the weight of that node. Suppose that the given number is 24, then there exists 4 different paths which have the same given weight: {10 5 2 7}, {10 4 10}, {10 3 3 6 2} and {10 3 3 6 2}, which correspond to the red edges in Figure 1.

Figure 1

Input Specification:

Each input file contains one test case. Each case starts with a line containing 0 < N <= 100, the number of nodes in a tree, M (< N), the number of non-leaf nodes, and 0 < S < 2³⁰, the given weight number. The next line contains N positive numbers where W_i (<1000) corresponds to the tree node T_i. Then M lines follow, each in the format:

ID K ID[1] ID[2] ... ID[K]

where ID is a two-digit number representing a given non-leaf node, K is the number of its children, followed by a sequence of two-digit ID's of its children. For the sake of simplicity, let us fix the root ID to be 00.

Output Specification:

For each test case, print all the paths with weight S in non-increasing order. Each path occupies a line with printed weights from the root to the leaf in order. All the numbers must be separated by a space with no extra space at the end of the line.

Note: sequence {A₁, A₂, ..., A_n} is said to be greater than sequence {B₁, B₂, ..., B_m} if there exists 1 <= k < min{n, m} such that A_i = B_ifor i=1, ... k, and A_k+1 > B_k+1.

Sample Input:

20 9 24
10 2 4 3 5 10 2 18 9 7 2 2 1 3 12 1 8 6 2 2
00 4 01 02 03 04
02 1 05
04 2 06 07
03 3 11 12 13
06 1 09
07 2 08 10
16 1 15
13 3 14 16 17
17 2 18 19

Sample Output:

提交代码

题目大意就不说了，写完后最后一个点始终段错误，开始以为数组开小了，结果不是，是cmp函数的问题，函数里最后的判断我写的是如果两条路径相同则返回true，结果这样会段错误，改成返回false则过了，不是很理解（不过想了想，相等的情况应该是返回false的）。ps：之后在网上搜了搜，发现了其中的原因，这里把网站附上点击打开链接

#include<bits/stdc++.h>
using namespace std;
vector<int>a[105],path[105];
int v[105],s,l,temp[1500005];
void dfs(int root,int pos,int sum){
     temp[pos]=v[root];
     sum+=temp[pos];
     //cout<<temp[pos]<<" "<<root<<" "<<pos<<endl;
     int len=a[root].size();
     if((len==0&&sum!=s)||sum>s||(sum==s&&len!=0)) return;
     if(sum==s&&len==0) {
        /*for(int i=1;i<=pos;i++)
            printf("%d ",temp[i]);
        printf("\n");*/
        for(int i=1;i<=pos;i++)
        path[l].push_back(temp[i]);
        l++;
        return;
     }
     for(int i=0;i<len;i++){
         dfs(a[root][i],pos+1,sum);
     }
}
bool cmp(int x,int y){
    int len=min(path[x].size(),path[y].size());
    for(int i=0;i<len;i++){
        if(path[x][i]>path[y][i]) return true;
        if(path[x][i]<path[y][i]) return false;

    }
    return false;
}


int main(){
    int n,m,k,x,y;
    l=0;
    scanf("%d%d%d",&n,&m,&s);
    for(int i=0;i<n;i++){
        scanf("%d",&v[i]);
    }
    while(m--){
        scanf("%d%d",&x,&k);
        for(int i=1;i<=k;i++){
            scanf("%d",&y);
            a[x].push_back(y);
        }
    }
    dfs(0,1,0);
    int p[1005];
    for(int i=0;i<l;i++)p[i]=i;
    sort(p,p+l,cmp);
    for(int i=0;i<l;i++){
        int len=path[p[i]].size();
        for(int j=0;j<len;j++){
            printf("%d",path[p[i]][j]);
            if(j!=len-1) printf(" ");
        }
        printf("\n");
    }

    return 0;
}