Submission #1195045

#	Time	Username	Problem	Language	Result	Execution time	Memory
1195045		nikulid	이상적인 도시 (IOI12_city)	C++20	0 / 100	150 ms	327680 KiB

city

#include <climits>
#include <vector>
#include <queue>

/*

  lemma: I will solve subtasks 1 and 2
  proof:

*/
using namespace std;

queue<vector<int>> q;

long long ans=0;
vector<vector<int>> visitted, exists; // both are [y][x]
vector<vector<bool>> DONE;
void bfs(int I, int x, int y, int curdist, int oy, int ox){
  
  ans %= 1000000000;
  vector<int> nxt(6);

  curdist++;
  
  nxt[0]=I;
  nxt[3]=curdist;
  nxt[4]=oy;
  nxt[5]=ox;
  
  // coords is [x,y]
  visitted[y][x] = I;
  if(exists[y+1][x+1] && visitted[y+1][x+1] != I){
    if(y >= oy && x >= ox){
      ans += curdist;
      //bfs(I, x+1, y+1, curdist, oy, ox);
    }
    nxt[1]=x+1;
    nxt[2]=y+1;
    q.push(nxt);
  }
  if(exists[y+1][x-1] && visitted[y+1][x-1] != I){
    if(y >= oy && x >= ox){
      ans += curdist;
      //bfs(I, x-1, y+1, curdist, oy, ox);
    }
    nxt[1]=x-1;
    nxt[2]=y+1;
    q.push(nxt);
  }
  if(exists[y-1][x+1] && visitted[y-1][x+1] != I){
    if(y >= oy && x >= ox){
      ans += curdist;
      //bfs(I, x+1, y-1, curdist, oy, ox);
    }
    nxt[1]=x+1;
    nxt[2]=y-1;
    q.push(nxt);
  }
  if(exists[y-1][x-1] && visitted[y-1][x-1] != I){
    if(y >= oy && x >= ox){
      ans += curdist;
      //bfs(I, x-1, y-1, curdist, oy, ox);
    }
    nxt[1]=x-1;
    nxt[2]=y-1;
    q.push(nxt);
  }
  return;
}

int DistanceSum(int N, int *X, int *Y) {
  DONE = vector<vector<bool>>(N+1, vector<bool>(N+1, 0));
  visitted = vector<vector<int>>(N+1, vector<int>(N+1, 0));
  exists = vector<vector<int>>(N+1, vector<int>(N+1, 0));

  // find min X and Y values, to translate the graph to start at (1,1)
  int minX=INT_MAX-1, minY=INT_MAX-1;
  for(int i=0; i<N; i++){
    minX = min(minX, X[i]);
    minY = min(minY, Y[i]);
  }
  for(int i=0; i<N; i++){
    exists[Y[i]-minY][X[i]-minX] = 1;
  }
  vector<int> cur;
  for(int i=0; i<N; i++){
    bfs(i, X[i]-minX, Y[i]-minY, 0, X[i]-minX, Y[i]-minY);
    while(!q.empty()){
      cur = q.front();
      q.pop();
      bfs(cur[0], cur[1], cur[2], cur[3], cur[4], cur[5]);
    }
  }
  ans %= 1000000000;
  return ans;
}

• Subtask 1: 0 / 11
• Subtask 2: 0 / 21
• Subtask 3: 0 / 23
• Subtask 4: 0 / 45