`[[0, 1, 10], [2, 0, 5]]`

.
Given a list of transactions between a group of people, return the minimum number of transactions required to settle the debt.

**Note:**

- A transaction will be given as a tuple (x, y, z). Note that
`x ? y`

and`z > 0`

. - Person's IDs may not be linear, e.g. we could have the persons 0, 1, 2 or we could also have the persons 0, 2, 6.

**Example 1:**

Input:[[0,1,10], [2,0,5]]Output:2Explanation:Person #0 gave person #1 $10. Person #2 gave person #0 $5. Two transactions are needed. One way to settle the debt is person #1 pays person #0 and #2 $5 each.

**Example 2:**

Input:[[0,1,10], [1,0,1], [1,2,5], [2,0,5]]Output:1Explanation:Person #0 gave person #1 $10. Person #1 gave person #0 $1. Person #1 gave person #2 $5. Person #2 gave person #0 $5. Therefore, person #1 only need to give person #0 $4, and all debt is settled.

**[Thoughts]**

很有意思的一道题。首先，对图进行处理，算清楚每个人的负债， 有两个前提：

1. 最后的负债一定可以清零

2. 题目不要求保留原有付款关系。

所以，对图做个dfs即可。

**[Code]**

```
1: int minTransfers(vector<vector<int>>& trans) {
2: unordered_map<int, long> bal; // balance on each person
3: for(auto t: trans) {
4: bal[t[0]] -= t[2];
5: bal[t[1]] += t[2];
6: }
7:
8: vector<long> debt;
9: for(auto b: bal) {
10: // only track the person who has debt
11: if(b.second) debt.push_back(b.second);
12: }
13: return dfs(0, 0, debt);
14: }
15:
16: // get the min number of transactions starting from s
17: int dfs(int s, int cnt, vector<long>& debt) {
18: while (s < debt.size() && !debt[s]) ++s; // skip all zero debt
19:
20: int res = INT_MAX;
21: for (long i = s+1, prev = 0; i < debt.size(); ++i) {
22: // skip same value or same sign debt
23: if (debt[i] != prev && debt[i]*debt[s] < 0){
24: debt[i] += debt[s];
25: res = min(res, dfs(s+1,cnt+1, debt));
26: debt[i]-=debt[s];
27: prev = debt[i];
28: }
29: }
30: return res < INT_MAX? res : cnt;
31: }
```

## No comments:

Post a Comment