HDU 3661 Assignments (水题，贪心)

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;
}