1053 Path of Equal Weight (30)（30 分）

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 < 2^30^, the given weight number. The next line contains N positive numbers where W~i~ (&lt1000) 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 {A~1~, A~2~, ..., A~n~} is said to be greater than sequence {B~1~, B~2~, ..., B~m~} if there exists 1 <= k < min{n, m} such that A~i~ = B~i~ for 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:

10 5 2 7
10 4 10
10 3 3 6 2
10 3 3 6 2


排序是按权值，不是结点编号。
代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
#include <map>
using namespace std;
int n,m,s,id,k,d;
int w[101],f[101],sw[101],noleaf[101],p;
vector<int> path[101];
void getpath(int t) {///记录到叶节点的路径 和 总权值和
    if(f[t] != -1) {///父节点存在
        getpath(f[t]);
        sw[t] = w[t] + sw[f[t]];
    }
    else sw[t] = w[t];
    path[p].push_back(w[t]);
}
int main() {
    scanf("%d%d%d",&n,&m,&s);
    for(int i = 0;i < n;i ++) {
        scanf("%d",&w[i]);
        f[i] = -1;
    }
    for(int i = 0;i < m;i ++) {
        scanf("%d%d",&id,&k);
        noleaf[id] = 1;
        for(int j = 0;j < k;j ++) {
            scanf("%d",&d);
            f[d] = id;
        }
    }
    for(int i = 0;i < n;i ++) {
        if(noleaf[i])continue;
        getpath(i);
        if(sw[i] == s) {
            p ++;
        }
        else {
            path[p].clear();
        }
    }
    sort(path,path + p);
    for(int i = p - 1;i >= 0;i --) {
        for(int j = 0;j < path[i].size();j ++) {
            if(j)putchar(' ');
            printf("%d",path[i][j]);
        }
        putchar('\n');
    }
}