Codechef Chef and Triangles(离散化+区间并集)

题目链接 Chef and Triangles

先排序，然后得到$m - 1$个区间:

$(a[2] - a[1], a[2] ＋ a[1])$

$(a[3] - a[2], a[3] ＋ a[2])$

$……$

$(a[n] - a[n - 1], a[n] + a[n - 1])$

对这些区间求交集  再和$[L, R]$求并集，最后的元素个数就是答案。

 #include <bits/stdc++.h>

 using namespace std;

 #define rep(i,a,b)              for(int i(a); i <= (b); ++i)
 #define LL              long long
 #define INF            1 << 30

 const int N =  + ;

 struct node{ LL x, y;} c[N], q[N];

 struct Node{
     LL x;  int y;
     friend bool operator < (const Node &a, const Node &b){
         return (a.x == b.x) ? a.y < b.y : a.x < b.x;
     }
 } p[N];

 map <LL, int> mp;
 LL a[N], ori[N], fp[N], l, r, x, y, ans;
 int n, cnt, et, nx, ny, now;

 int f[N], h[N], d[N], ret, cnt_status, _min, _max;

 int main(){

     scanf("%d%lld%lld", &n, &l, &r);
     rep(i, , n) scanf("%lld", a + i);
     sort(a + , a + n + );

     cnt = ; et = ;
     rep(i, , n - ){
         c[++cnt].x = a[i + ] - a[i] + ;
         p[++et].x = c[cnt].x;     p[et].y = et;
         c[cnt].y = a[i + ] + a[i] - ;
         p[++et].x = c[cnt].y;   p[et].y = et;
     }

     rep(i, , et) ori[i] = p[i].x;
     sort(p + , p + et + );

     f[p[].y] = ; rep(i, , et) f[p[i].y] = p[i].x == p[i - ].x ? f[p[i - ].y] : f[p[i - ].y] + ;

     rep(i, , et) f[i] *= ;

     rep(i, , et){
         mp[ori[i]] = f[i];
         fp[f[i]] = ori[i];
     }

     _min = INF;
     _max = -_min;

     memset(h, , sizeof h);
     rep(i, , cnt){
         x = c[i].x, y = c[i].y;
         nx = mp[x], ny = mp[y];

         _min = min(_min, nx);
         _max = max(_max, ny + );

         ++h[nx], --h[ny + ];
     }
     now = ;
     rep(i, _min, _max){
         now += h[i];
         d[i] = now;
     }

     cnt_status = ;
     ret = ;
     rep(i, _min, _max){
         if (cnt_status ==  && d[i]){
             ++ret;
             q[ret].x = fp[i];
             cnt_status = ;
         }

         else
         if (cnt_status ==  && d[i] == ){
             q[ret].y = fp[i - ];
             cnt_status = ;
         }
     }

     ans = ;
     rep(i, , ret){
         x = max(l, q[i].x), y = min(r, q[i].y);
         if (x <= y) ans += y - x + ;
     }

     printf("%lld\n", ans);

     return ;

 }