题意:n个工人,有n件工作a,n件工作b,每个工人干一件a和一件b,a[i] ,b[i]代表工作时间,如果a[i]+b[j]>t,则老板要额外付钱a[i]+b[j]-t;现在要求老板付钱最少;
析:贪心策略,让大的和小的搭配,小的和大的搭配,是最优的。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <stack> using namespace std; typedef long long LL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const double inf = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 1000 + 5; const int mod = 1e9; const char *mark = "+-*"; const int dr[] = {-1, 0, 1, 0}; const int dc[] = {0, 1, 0, -1}; int n, m; inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } int a[maxn], b[maxn]; bool cmp(const int &lhs, const int &rhs){ return lhs > rhs; } int main(){ while(scanf("%d %d", &n, &m) == 2){ for(int i = 0; i < n; ++i) scanf("%d", &a[i]); for(int j = 0; j < n; ++j) scanf("%d", &b[j]); sort(a, a+n); sort(b, b+n, cmp); int ans = 0; for(int i = 0; i < n; ++i) ans += a[i] + b[i] - m > 0 ? a[i] + b[i] - m : 0; printf("%d\n", ans); } return 0; }
时间: 2024-11-05 18:41:41