Submission #501953

#	Time	Username	Problem	Language	Result	Execution time	Memory
501953		ETK	Dungeon 2 (JOI16_dungeon2)	C++14	0 / 100	11 ms	2212 KiB

dungeon2

#include "dungeon2.h"
#include <bits/stdc++.h>
#define rep(i,a,b) for(int i = (a);i <= (b);++i)
using namespace std;
int d[205],T[205][205],cnt = 1,dis[205][205],ans[205];
//1: Edges on tree, 3: Reverse of 1, 2: nontree edges
int type[205][205];
void dfs(int x){
    d[x] = NumberOfRoads();
    rep(i,1,d[x]){
        Move(i,2);
        int col = Color(),lst = LastRoad();
        type[x][i] = col;
        if(col == 1){//not visited yet
            T[x][i] = ++cnt;
            dfs(cnt);
            type[T[x][i]][lst] = 3;
            T[T[x][i]][lst] = -1;
        }else if(col == 2){
            T[x][i] = 0;
        }
        //Going back
        if(col == 1)Move(lst,3);
        else Move(lst,col);
    }
}
//用三进制分解找到非树边的编号
void work(int x,int pow3){
    rep(i,1,d[x]){
        if(type[x][i] != 3){
            Move(i,x/pow3%3+1);
            int lst = LastRoad();
            if(type[x][i] == 1)work(T[x][i],pow3);
            else T[x][i] += pow3 * (Color() - 1);
            Move(lst,Color());
        }
    }
}
void Inspect(int R){
    memset(T,-1,sizeof(T));
    dfs(1);
    int now = 1;
    rep(i,0,4){
        work(1,now);
        now *= 3;
    }
    memset(dis,0x3f,sizeof(dis));
    rep(i,1,cnt)rep(j,1,d[i])
        dis[i][T[i][j]] = 1;
    rep(i,1,cnt)rep(j,1,cnt)rep(k,1,cnt){
        dis[i][j] = min(dis[i][j],dis[i][k] + dis[k][j]);
    }
    rep(i,1,cnt)rep(j,i+1,cnt)
        ans[dis[i][j]]++;
    rep(i,1,R) Answer(i,ans[i]);
}

• Subtask 1: 0 / 17
• Subtask 2: 0 / 27
• Subtask 3: 0 / 56