Given a non-empty tree with root R, and with weight Wi assigned to each tree node Ti. 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
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 Wi (<1000) corresponds to the tree node Ti. Then M lines follow, each in the format:
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1 |
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, …, An} is said to be greater than sequence {B1, B2, …, Bm} if there exists 1 <= k < min{n, m} such that Ai = Bi for i=1, … k, and Ak+1 > Bk+1.
Sample Input:
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1 2 3 4 5 6 7 8 9 10 11 |
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:
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1 2 3 4 |
10 5 2 7 10 4 10 10 3 3 6 2 10 3 3 6 2 |
代码如下：
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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 |
#include<stdio.h> #include<stdlib.h> typedef struct info_ { int parent; int is_leaf; int weight; } info; info node[100]; int compare(const void *a,const void *b) { int *pa,*pb; pa=*(int **)a; pb=*(int **)b; while(*pa!=-1&&*pb!=-1) { if(*pa>*pb) { return -1; } else if(*pa<*pb) { return 1; } pa++; pb++; } if(*pa==-1) { return 1; } return -1; } int main() { int n,i,j,m,k,s,temp,sum,parent,child,record[100],*result[100],total,offset=0; scanf("%d%d%d",&n,&m,&s); for(i=0;i<n;i++) { scanf("%d",&(node[i].weight)); node[i].is_leaf=1; } for(i=0;i<m;i++) { scanf("%d%d",&parent,&k); node[parent].is_leaf=0; for(j=0;j<k;j++) { scanf("%d",&child); node[child].parent=parent; } } for(i=0;i<n;i++) { if(node[i].is_leaf) { temp=i; sum=0; while(temp!=0) { sum+=node[temp].weight; temp=node[temp].parent; } sum+=node[temp].weight; if(sum==s) { record[offset]=i; result[offset]=(int *)malloc(n*sizeof(int)); temp=i; j=0; while(temp!=0) { result[offset][j++]=node[temp].weight; temp=node[temp].parent; } result[offset][j++]=node[temp].weight; result[offset][j]=-1; total=j; for(j=0;j<total/2;j++) { temp=result[offset][j]; result[offset][j]=result[offset][total-j-1]; result[offset][total-j-1]=temp; } offset++; } } } qsort(result,offset,sizeof(int *),compare); for(i=0;i<offset;i++) { printf("%d",result[i][0]); j=1; while(result[i][j]!=-1) { printf(" %d",result[i][j++]); } printf("\n"); } } |