题意:给定两个数组 构造一个配对 使得每个位置上之差的和最小 大小相同的不能配对

题解:猜了一下 应该把两个数组排序一下 答案最小

　　 处理相同的不能配对的时候 那么应该找就近的交换

　　 猜一下 如果有合法方案的话 肯定一个大小为3的环就能解决配对问题了

#include 
      
       using namespace std;typedef long long ll;const ll INF = 1e12; int q[100005];int w[100005];ll dp[100005]; ll abss(ll x, ll y){    if(x == y) return INF;    else if(x > y) return x - y;    else if(x < y) return y - x;} int main(){    dp[0] = 0;    int n;    scanf("%d", &n);    for(int i = 1; i <= n; i++) scanf("%d%d", &q[i], &w[i]), dp[i] = INF;    sort(q + 1, q + 1 + n);    sort(w + 1, w + 1 + n);     for(int i = 1; i <= n; i++)    {        if(i >= 1 && dp[i - 1] != INF)        {            //if(i == 3) cout<<"hi"<
       
        = 2 && dp[i - 2] != INF)        {            ll x1 = abss(q[i], w[i - 1]);            ll x2 = abss(q[i - 1], w[i]);            dp[i] = min(dp[i], dp[i - 2] + x1 + x2);            //if(i == 3) cout<
        
         <
         
          = 3 && dp[i - 3] != INF)        {            ll x1 = abss(q[i], w[i - 2]);            ll y1 = abss(q[i - 1], w[i - 1]);            ll y2 = abss(q[i - 2], w[i]);            dp[i] = min(dp[i], dp[i - 3] + y1 + y2 + x1);             ll z1 = abss(q[i - 1], w[i]);            ll z2 = abss(q[i - 2], w[i - 1]);            dp[i] = min(dp[i], dp[i - 3] + z1 + z2 + x1);             ll o1 = abss(q[i], w[i - 1]);            ll o2 = abss(q[i - 1], w[i - 2]);            ll o3 = abss(q[i - 2], w[i]);            dp[i] = min(dp[i], dp[i - 3] + o1 + o2 + o3);            //if(i == 3) cout<
          
           <