HDU 4941 Magical Forest 离散化

题意： 一个n * m 的图， 某些点有苹果，每个苹果有价值， 然后现在交换某些行或者列， 或者查询某个点的苹果的价值（没有苹果就是0）。

还是比较好想的。。x[],y[]数组存当前状态下的各行各列存的是初始时的哪一行哪一列，交换的时候交换值就行了， 可能复杂就在离散化吧，不过实际上用map离散化也挺easy的样子。。虽然我用的最传统的vector、sort、 unique、resize然后upper_bound、、

题目给了12s实际上3s完全够了。。

#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cmath>
#include <string>
#include <cctype>
#include <cstdio>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <sstream>
#include <iostream>
#include <algorithm>
using namespace std;
#define ls id<<1,l,mid
#define rs id<<1|1,mid+1,r
#define OFF(x) memset(x,-1,sizeof x)
#define CLR(x) memset(x,0,sizeof x)
#define MEM(x) memset(x,0x3f,sizeof x)
typedef long long ll ;
typedef pair<int,int> pii ;
const int maxn = 3e5+50 ;
const int inf = 0x3f3f3f3f ;
const int MOD = 1e9+7 ;

struct pnt{
    int x, y ,c;
    pnt(int x, int y, int c):x(x),y(y),c(c){}
};

int N, M, n, m, k, q, T;
vector<pnt> fruit, query;
vector<int> Vx, Vy;

int mx[maxn], my[maxn];
map<pii, int> mp;
void init() {
    sort(Vx.begin(), Vx.end());
    sort(Vy.begin(), Vy.end());
    Vx.resize(unique(Vx.begin(), Vx.end()) - Vx.begin());
    Vy.resize(unique(Vy.begin(), Vy.end()) - Vy.begin());
    for (pnt e : fruit) {
        int x = e.x, y = e.y;
        x = upper_bound(Vx.begin(), Vx.end(), x) - Vx.begin();
        y = upper_bound(Vy.begin(), Vy.end(), y) - Vy.begin();
        mp[pii(x,y)] = e.c;
    }
    n = Vx.size();m = Vy.size();
}

inline int read() {
    char c = getchar();
    while (!isdigit(c)) c = getchar();
    int x = 0;
    while (isdigit(c)) {
        x = x * 10 + c - '0' ;
        c = getchar();
    }
    return x;
}

int main () {
#ifdef LOCAL
	freopen("C:\\Users\\Administrator\\Desktop\\in.txt","r",stdin);
//      freopen("C:\\Users\\Administrator\\Desktop\\out.txt","w",stdout);
#endif
    scanf("%d", &T);
    int cas = 1;
    while (T--) {
        printf("Case #%d:\n",cas++);
        scanf("%d%d%d", &n, &m, &k);
        Vx.clear();Vy.clear();
        fruit.clear();query.clear();
        mp.clear();
        Vx.push_back(n);Vy.push_back(m);
        Vx.push_back(0);Vy.push_back(0);
        int x, y, c;
        while (k--) {
            x = read(), y = read(), c = read();
            Vx.push_back(x);
            Vy.push_back(y);
            fruit.push_back(pnt(x,y,c));
        }
        q = read();
        while (q--) {
            c = read(), x = read(), y = read();
            if (c == 1) {
                Vx.push_back(x);
                Vx.push_back(y);
            } else if (c == 2) {
                Vy.push_back(x);
                Vy.push_back(y);
            } else {
                Vx.push_back(x);
                Vy.push_back(y);
            }
            query.push_back(pnt(x, y, c));
        }
        init();
        for (int i = 1; i <= n; i++) mx[i] = i;
        for (int i = 1; i <= m; i++) my[i] = i;
        for (pnt q : query) {
            int op = q.c;
            if (op == 3) {
                int x = upper_bound(Vx.begin(), Vx.end(), q.x) - Vx.begin();
                int y = upper_bound(Vy.begin(), Vy.end(), q.y) - Vy.begin();
                x = mx[x], y = my[y];
                if (mp.count(pii(x,y))) printf("%d\n",mp[pii(x,y)]);
                else puts("0");
            } else if (op == 1) {
                int x1 = upper_bound(Vx.begin(), Vx.end(), q.x) - Vx.begin();
                int x2 = upper_bound(Vx.begin(), Vx.end(), q.y) - Vx.begin();
                swap(mx[x1], mx[x2]);
            } else {
                int y1 = upper_bound(Vy.begin(), Vy.end(), q.x) - Vy.begin();
                int y2 = upper_bound(Vy.begin(), Vy.end(), q.y) - Vy.begin();
                swap(my[y1], my[y2]);
            }
        }
    }

	return 0;
}