UVaLive 6698 Sightseeing Bus Drivers (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。

#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;
}