【算法学习笔记】78. STL二分的练习 下标映射的处理技巧 SJTU OJ 1053 二哥的内存

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
using namespace std;

struct Point
{
    int x;
    int y;
    int z;
};

int n;
const int MAXN = 100000;
//int data[MAXN][MAXN];
Point points[MAXN];
int curx[MAXN];
int cury[MAXN];

void init(){
    cin>>n;
    for (int i = 0; i < n; ++i){
        //int x,y,z;
        //scanf("%d %d %d",&x,&y,&z);
        //data[x][y] = z;
        scanf("%d %d %d",&points[i].x,&points[i].y,&points[i].z);
    }
    
    for (int i = 0; i < MAXN; ++i)
        curx[i]=i;
    for (int i = 0; i < MAXN; ++i)
        cury[i]=i;
    
}

bool cmpPoint(const Point& a,const Point& b){
    if(a.x!=b.x){
        return a.x<b.x;
    }else
        return a.y<b.y;
}


int find(int x,int y){
    Point tofind;
    tofind.x = x;
    tofind.y = y;
    Point* f = lower_bound(points,points+n,tofind,cmpPoint);
    //注意lower_bound的返回值是大于等于val的第一个 不一定正好是它
    if(f != points+n and f->x==x and f->y==y)
        return f->z;
    return 0;
}

void exe(){
    int m;
    cin>>m;
    sort(points,points+n,cmpPoint);
    for (int i = 0; i < m; ++i)
    {
        int op,x,y;
        scanf("%d %d %d",&op,&x,&y);
        if(op==0){
            swap(curx[x],curx[y]);
        }else if(op==1)
            swap(cury[x],cury[y]);
        else{
            //printf("%d\n", data[curx[x]][cury[y]]);
            printf("%d\n", find(curx[x],cury[y]));
        }
    }
}


int main(int argc, char const *argv[])
{
    init();
    exe();
    return 0;
}