Problem link: http://codeforces.com/problemset/problem/463/C
To find 2 positions such that bishops in these positions cannot attach each other, we have to select 2 coordinates such that sum of x,y(row,column) in one coordinate is even and odd in other one.
Find the sum of primary and secondary diagonal and calculate the max amount of money for each position. After that iterate through each position and find the maximum money,position in even sum index set and odd sum index set.
This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters.
Learn more about bidirectional Unicode characters
| #include <vector> | |
| #include <list> | |
| #include <map> | |
| #include <set> | |
| #include <deque> | |
| #include <queue> | |
| #include <stack> | |
| #include <bitset> | |
| #include <algorithm> | |
| #include <functional> | |
| #include <numeric> | |
| #include <utility> | |
| #include <sstream> | |
| #include <iostream> | |
| #include <iomanip> | |
| #include <cstdio> | |
| #include <cmath> | |
| #include <cstdlib> | |
| #include <ctime> | |
| #include <cstring> | |
| #include <climits> | |
| #include <stdlib.h> | |
| #include <stdio.h> | |
| using namespace std; | |
| #define REP(i,n) for(int i=0; i<n; i++) | |
| #define FOR(i,st,end) for(int i=st;i<end;i++) | |
| #define db(x) cout << (#x) << " = " << x << endl; | |
| #define mp make_pair | |
| #define pb push_back | |
| #define MAX 100005 | |
| typedef long long int ll; | |
| ll secondary[4001]; | |
| ll primary[4001]; | |
| pair<int,int> evenIndex; | |
| pair<int,int> oddIndex; | |
| ll value[2]; | |
| int arr[2001][2001]; | |
| //value 0 denotes even | |
| //value 1 denotes odd | |
| int main(){ | |
| memset(value,-1,sizeof value); | |
| int n; | |
| cin>>n; | |
| REP(i,n){ | |
| REP(j,n){ | |
| scanf("%d",&arr[i][j]); | |
| secondary[i+j]+=arr[i][j]; | |
| primary[i-j+n-1]+=arr[i][j]; | |
| } | |
| } | |
| ll candidate; | |
| REP(i,n){ | |
| REP(j,n){ | |
| candidate=secondary[i+j]+primary[i-j+n-1]-arr[i][j]; | |
| if((i+j)&1){ | |
| if(candidate>value[1]){ | |
| value[1]=candidate; | |
| oddIndex.first=i; | |
| oddIndex.second=j; | |
| } | |
| } | |
| else{ | |
| if(candidate>value[0]){ | |
| value[0]=candidate; | |
| evenIndex.first=i; | |
| evenIndex.second=j; | |
| } | |
| } | |
| } | |
| } | |
| cout<<value[0]+value[1]<<endl; | |
| if(value[0]>value[1]){ | |
| cout<<evenIndex.first+1<<" "<<evenIndex.second+1<<" "; | |
| cout<<oddIndex.first+1<<" "<<oddIndex.second+1<<endl; | |
| } | |
| else{ | |
| cout<<oddIndex.first+1<<" "<<oddIndex.second+1<<" "; | |
| cout<<evenIndex.first+1<<" "<<evenIndex.second+1<<endl; | |
| } | |
| } |
coding hangover 