POJ 1852 Ants (等价思考)

题意：在一根杆上有 n 只蚂蚁，速度为1，方向不定，如果相碰，则反向运动，问你最长的时间和最短时间，所有蚂蚁都掉下杆去。

析：换个方法想，如果两只蚂蚁相碰了，会有什么现象？其实就和没有碰撞是一样的，没有区别，那么这个题就简单了，只要全都扫一遍即可。

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 <cmath>
#include <stack>
#include <list>
#include <sstream>
#define frer freopen("in.txt", "r", stdin)
#define frew 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 = 1e6 + 5;
const int mod = 1e9 + 7;
const int dr[] = {-1, 1, 0, 0};
const int dc[] = {0, 0, 1, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
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];

int main(){
    int T;  cin >> T;
    int l;
    while(T--){
        scanf("%d %d", &l, &n);
        for(int i = 0; i < n; ++i)  scanf("%d", &a[i]);
        int minans = 0, maxans = 0;
        for(int i = 0; i < n; ++i){
            minans = Max(minans, Min(a[i], l - a[i]));
            maxans = Max(maxans, Max(a[i], l - a[i]));
        }
        printf("%d %d
", minans, maxans);
    }
    return 0;
}