2020牛客寒假算法基础集训营4-I 匹配星星【贪心】

链接：https://ac.nowcoder.com/acm/contest/3005/I
来源：牛客网

示例1

输入

复制
2
1 1 0
2 2 1

2
1 1 0
2 2 1

输出

复制 1

1

示例2

输入

复制 2
1 1 1
2 2 1

2
1 1 1
2 2 1

输出

复制 0

0

备注:

思路：

　　

 #include <bits/stdc++.h>
 #define dbg(x) cout << #x << "=" << x << endl
 
 using namespace std;
 typedef long long LL;
 
 template<class T>inline void read(T &res)
 {
     char c;T flag=;
     while((c=getchar())<''||c>'')if(c=='-')flag=-;res=c-'';
     while((c=getchar())>=''&&c<='')res=res*+c-'';res*=flag;
 }
 
 namespace _buff {
     const size_t BUFF =  << ;
     char ibuf[BUFF], *ib = ibuf, *ie = ibuf;
     char getc() {
         if (ib == ie) {
             ib = ibuf;
             ie = ibuf + fread(ibuf, , BUFF, stdin);
         }
         return ib == ie ? - : *ib++;
     }
 }
 
 int qread() {
     using namespace _buff;
     int ret = ;
     bool pos = true;
     char c = getc();
     for (; (c < '' || c > '') && c != '-'; c = getc()) {
         assert(~c);
     }
     if (c == '-') {
         pos = false;
         c = getc();
     }
     for (; c >= '' && c <= ''; c = getc()) {
         ret = (ret << ) + (ret << ) + (c ^ );
     }
     return pos ? ret : -ret;
 }
 
 const int maxn = 1e5 + ;
 
 set <int> s;
 map<int, int> mapp;
 
 int n;
 
 struct node {
     int x,y,z;
 }a[maxn];
 
 bool cmp(node a, node b) {
     if(a.x == b.x) {
         return a.z > b.z;
     }
     return a.x < b.x;
 }
 
 int main()
 {
     read(n);
     for(int i = ; i <= n; ++i) {
         scanf("%d %d %d",&a[i].x, &a[i].y, &a[i].z);
     }
     sort(a+, a+n+, cmp);
     set<int>::iterator it;
 
     LL ans = ;
     for(int i = ; i <= n; ++i) {
         if(a[i].z == ) {
             if(!mapp[a[i].y]) {
                 s.insert(a[i].y);
             }
             mapp[a[i].y]++;
         }
         if(a[i].z == ) {
             it = s.lower_bound(a[i].y);
             if(it != s.begin()) {
                 mapp[*it]--;
                 if(!mapp[*it]) {
                     s.erase(*it);
                 }
                 dbg(*it);
                 ans++;
             }
         }
     }
     cout << ans << endl;
     return ;
 }
 /*
 5
 1 1 0
 2 2 0
 3 3 0
 4 4 0
 5 5 1
 */