[C++知识库]codeforces 775-div2 C(摆大烂)

题目
题意: 给定n*m的矩阵，矩阵每个位置有对应的颜色。求同种颜色任意两对坐标之间的曼哈顿距离之和。dis(x1,y1,x2,y2) = abs(x1 - x2) + abs(y1 - y2)
思路: x和y无关，可以分开统计贡献。把颜色col对应的所有点的x、y坐标分别存起来，之后sort一下。

时间复杂度: (tlogt, t = nm)
代码:

// Problem: C. Weird Sum
// Contest: Codeforces - Codeforces Round #775 (Div. 2, based on Moscow Open Olympiad in Informatics)
// URL: https://codeforces.com/contest/1649/problem/C
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<complex>
#include<cstring>
#include<cmath>
#include<vector>
#include<map>
#include<unordered_map>
#include<list>
#include<set>
#include<queue>
#include<stack>
#define OldTomato ios::sync_with_stdio(false),cin.tie(nullptr),cout.tie(nullptr)
#define fir(i,a,b) for(int i=a;i<=b;++i) 
#define mem(a,x) memset(a,x,sizeof(a))
#define p_ priority_queue
// round() 四舍五入 ceil() 向上取整 floor() 向下取整
// lower_bound(a.begin(),a.end(),tmp,greater<ll>()) 第一个小于等于的
// #define int long long //QAQ
using namespace std;
typedef complex<double> CP;
typedef pair<int,int> PII;
typedef long long ll;
// typedef __int128 it;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const ll inf = 1e18;
const int N = 2e5+10;
const int M = 1e6+10;
const int mod = 1e9+7;
const double eps = 1e-6;
inline int lowbit(int x){ return x&(-x);}
template<typename T>void write(T x)
{
    if(x<0)
    {
        putchar('-');
        x=-x;
    }
    if(x>9)
    {
        write(x/10);
    }
    putchar(x%10+'0');
}
template<typename T> void read(T &x)
{
    x = 0;char ch = getchar();ll f = 1;
    while(!isdigit(ch)){if(ch == '-')f*=-1;ch=getchar();}
    while(isdigit(ch)){x = x*10+ch-48;ch=getchar();}x*=f;
}
int n,m,k,T;
vector<int> l[N],r[N];
ll suml[N],sumr[N];
void solve()
{
   read(n); read(m);
   for(int i=1;i<=n;++i)
   {
   	  for(int j=1;j<=m;++j)
   	  {
   	  	 int x; read(x);
   	  	 l[x].push_back(i);
   	  	 r[x].push_back(j);
   	  	 suml[x] += i;
   	  	 sumr[x] += j;
   	  }
   }
   ll ans = 0;
   for(int i=1;i<=(int)1e5;++i)
   {
   	   if(!l[i].size()) continue;
   	   sort(l[i].begin(),l[i].end());
   	   sort(r[i].begin(),r[i].end());
   	   int tot;
   	   tot = l[i].size() - 1;
   	   for(int j=0;j<l[i].size();++j)
   	   {
   	   	  int t = l[i][j];
   	   	  suml[i] -= t;
   	   	  ans += suml[i] - 1ll*tot * t; tot--;  
   	   }
   	   tot = r[i].size() - 1;
   	   for(int j=0;j<r[i].size();++j)
   	   {
   	   	  int t = r[i][j];
   	   	  sumr[i] -= t;
   	   	  ans += sumr[i] - 1ll*tot * t; tot--;  
   	   }
   }
   write(ans);
}
signed main(void)
{  
   T = 1;
   // OldTomato; cin>>T;
   // read(T);
   while(T--)
   {
   	 solve();
   }
   return 0;
}