46 lines
1.3 KiB
C++
46 lines
1.3 KiB
C++
#include<iostream>
|
||
#include<queue>
|
||
#include<cmath>
|
||
|
||
using namespace std;
|
||
|
||
struct hs //指花生:)
|
||
{
|
||
int x, y, v;
|
||
bool operator < (const hs & a) const
|
||
{
|
||
return v < a.v;
|
||
}//优先队列默认从大到小排序,但是需要用到小于号,所以这里重载运算符 < (只重载<就可以了)
|
||
};
|
||
|
||
priority_queue <hs> q;
|
||
|
||
int main()
|
||
{
|
||
int m, n, k;
|
||
int sum = 0;
|
||
cin >> m >> n >> k;
|
||
int a;
|
||
hs t;
|
||
for(int i = 1; i <= m; i++)
|
||
for(int j = 1; j <= n; j++)
|
||
{
|
||
cin >> a;
|
||
t.v = a;
|
||
t.x = j;
|
||
t.y = i;
|
||
if(a != 0) //如果有数值,就推进队列
|
||
q.push(t);
|
||
}
|
||
k -= q.top().y; //走到最大的花生植株
|
||
while(!q.empty() && k > q.top().y) //第一个是检验队列不为空,如果空了说明全摘完了,第二个是检验剩余的时间可以走回路边
|
||
{
|
||
k--; //只要进入一次循环,说明到了一棵植株,采摘耗费1时间
|
||
sum += q.top().v;
|
||
int lx = q.top().x, ly = q.top().y;
|
||
q.pop();//换下一棵树
|
||
k -= abs(lx - q.top().x) + abs(ly - q.top().y);//剩余的时间减去从上一棵植株到这棵的时间
|
||
}
|
||
cout << sum << endl;
|
||
return 0;
|
||
} |