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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。
#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 <cmath> #include <stack> #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) 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 = 1e5 + 5; const int mod = 1e8; const char *mark = "+-*"; const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1}; const int dc[] = {0, 1, 0, -1, -1, 1, 1, -1}; int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline int Min(int a, int b){ return a < b ? a : b; } inline int Max(int a, int b){ return a > b ? a : b; } inline LL Min(LL a, LL b){ return a < b ? a : b; } inline LL Max(LL a, LL b){ return a > b ? a : b; } inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } int a[maxn], b[maxn]; int main(){ int T; cin >> T; while(T--){ scanf("%d %d", &n, &m); for(int i = 0; i < n; ++i) scanf("%d", &a[i]); for(int i = 0; i < n; ++i) scanf("%d", &b[i]); sort(a, a+n); sort(b, b+n); int ans = 0; for(int i = 0, j = n-1; i < n; ++i, --j) ans += Max(0, a[i]+b[j]-m); printf("%d\n", ans); } return 0; }
UVaLive 6698 Sightseeing Bus Drivers (水题,贪心)
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原文地址:http://www.cnblogs.com/dwtfukgv/p/5801272.html