A - Warehouse Inventory Management
【题目来源】
AtCoder：A - Warehouse Inventory Management
【题目描述】
Takahashi works in the logistics department of a company. The company has \(N\) warehouses, numbered from warehouse \(1\) to warehouse \(N\). Each warehouse \(i\) \((1 \leq i \leq N)\) currently stores \(P_i\) items.
高桥在一家公司的物流部门工作。该公司有 \(N\) 个仓库，编号从 \(1\) 到 \(N\)。每个仓库 \(i\)（\(1 \leq i \leq N\)）目前存储了 \(P_i\) 件货物。
Today, \(M\) shipping instructions to move items between warehouses have been issued. The \(j\)-th \((1 \leq j \leq M)\) shipping instruction is "move \(W_j\) items from warehouse \(U_j\) to warehouse \(V_j\)." Here, \(U_j \neq V_j\). There may be multiple shipping instructions between the same pair of warehouses.
今天，下达了 \(M\) 条在仓库间移动货物的指令。第 \(j\) 条（\(1 \leq j \leq M\)）指令是"从仓库 \(U_j\) 移动 \(W_j\) 件货物到仓库 \(V_j\)"。这里，\(U_j \neq V_j\)。同一对仓库之间可能有多条移动指令。
The \(M\) shipping instructions are applied all at once, not sequentially. That is, the final number of items in each warehouse is obtained by taking the initial number of items, adding the total number of items coming into that warehouse, and subtracting the total number of items going out of that warehouse.
这 \(M\) 条指令是同时生效的，而不是顺序执行的。也就是说，每个仓库的最终货物数量是通过初始货物数量加上进入该仓库的货物总数，再减去从该仓库运出的货物总数得到的。
Takahashi wants to find the warehouse number with the most items after all shipping instructions have been applied. If there are multiple warehouses with the most items, find the one with the smallest warehouse number.
高桥希望找出在所有指令生效后，货物数量最多的仓库编号。如果有多个仓库的货物数量同为最多，则输出编号最小的仓库编号。
【输入】
\(N\) \(M\)
\(P_1\) \(P_2\) \(\ldots\) \(P_N\)
\(U_1\) \(V_1\) \(W_1\)
\(U_2\) \(V_2\) \(W_2\)
\(\vdots\)
\(U_M\) \(V_M\) \(W_M\)
- The first line contains the number of warehouses \(N\) and the number of shipping instructions \(M\), separated by a space.
- The second line contains the initial number of items in each warehouse \(P_1, P_2, \ldots, P_N\), separated by spaces.
- The following \(M\) lines contain the details of each shipping instruction. If \(M = 0\), this part is not given.
- The \(j\)-th line \((1 \leq j \leq M)\) contains the source warehouse number \(U_j\), the destination warehouse number \(V_j\), and the number of items to move \(W_j\) for the \(j\)-th shipping instruction, separated by spaces.
【输出】
Print on one line the warehouse number with the most items after all shipping instructions have been applied. If there are multiple warehouses with the most items, print the one with the smallest number.
【输入样例】
3 2
10 20 30
1 2 5
3 2 10
【输出样例】
2
【代码详解】
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N = 200005;
int n, m, p[N]; // n: 人数, m: 交易次数, p: 每个人的净收入
struct Node
{int u, v, w;
} a[N];
signed main()
{cin >> n >> m; // 输入人数和交易次数for (int i = 1; i <= n; i++) // 输入初始金额{cin >> p[i];}for (int i = 1; i <= m; i++) // 处理交易{int u, v, w;cin >> u >> v >> w; // 从u给v转账wp[u] -= w; // u减少wp[v] += w; // v增加w}// 调试输出// for (int i=1; i<=n; i++)// cout << p[i] << " ";// cout << endl;int maxn = -1, maxi = -1; // 最大净收入和对应的人for (int i = 1; i <= n; i++) // 查找最大净收入{if (p[i] > maxn){maxn = p[i];maxi = i;}}cout << maxi << endl; // 输出净收入最大的人的编号return 0;
}
【运行结果】
3 2
10 20 30
1 2 5
3 2 10
2