Codeforces Round #492 (Div. 2) [Thanks, uDebug!]

这次的题好奇怪哦。。。

C - Tesla

思路：先把跟停车位相邻的车停进去，然后开始转圈。。。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define pii pair<int, int>

using namespace std;

const int N = 20007;
const int M = 1e6 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 +7;

int a[4][N], n, k;
vector<int> ans[3];

pii getPos(int id) {
    if (id < n) return mk(1, id);
    else return mk(2, 2 * n - 1 - id);
}

void add(int x, int y, int z) {
    ans[0].push_back(x);
    ans[1].push_back(y);
    ans[2].push_back(z);
}

int main() {
    scanf("%d%d", &n, &k);

    for(int i = 0; i < 4; i++) {
        for(int j = 0; j < n; j++) {
            scanf("%d", &a[i][j]);
        }
    }

    for(int j = 0; j < n; j++) {
        if(a[0][j] == 0) continue;
        if(a[0][j] == a[1][j]) {
            add(a[0][j], 0, j);
            k--;
            a[1][j] = 0;
        }
    }

    for(int j = 0; j < n; j++) {
        if(a[3][j] == 0) continue;
        if(a[3][j] == a[2][j]) {
            add(a[3][j], 3, j);
            k--;
            a[2][j] = 0;
        }
    }

    if(k == 2 * n) {
        puts("-1");
        return 0;
    }

    while(k > 0) {
        for(int i = 0; i < 2 * n; i++) {
            pii u = getPos(i);
            if(a[u.first][u.second] == 0) continue;
            if(a[u.first ^ 1][u.second] == a[u.first][u.second]) {
                add(a[u.first][u.second], u.first ^ 1, u.second);
                k--;
                a[u.first][u.second] = 0;
                continue;
            }

            pii v = getPos((i + 2 * n - 1) % (2 * n));
            if(a[v.first][v.second] != 0) continue;
            add(a[u.first][u.second], v.first, v.second);
            swap(a[u.first][u.second], a[v.first][v.second]);
        }
    }

    printf("%d
", ans[0].size());
    for(int i = 0; i < ans[0].size(); i++) {
        printf("%d %d %d
", ans[0][i], ans[1][i] + 1, ans[2][i] + 1);
    }

    return 0;
}

E：

题目大意：给你n个长度小于1e6的向量，让你把每个向量变成正向或者负向，使得所有向量加起来的长度不超过1.5 * 1e6

思路：正解是每三个向量中都能找到两个向量合起来 <= 1e6，然后不断合并，最后只会剩下一个或者两个向量，

如果一个向量肯定 <= 1e6， 如果是两个向量一定 <= 1.5 * 1e6。。。

但是可以用随机洗牌的方法去贪心，因为能卡掉的数据特别不好造。。。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define pii pair<int, int>

using namespace std;

const int N = 1e5 + 7;
const int M = 1e6 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 +7;

int n, ans[N], id[N];
LL base = 1500000ll * 1500000;
struct Vector {
    Vector(LL x = 0, LL y = 0) {
        this->x = x;
        this->y = y;
    }

    Vector operator + (const Vector &rhs) const {
        return Vector(x + rhs.x, y + rhs.y);
    }

    Vector operator - (const Vector &rhs) const {
        return Vector(x - rhs.x, y - rhs.y);
    }

    LL len() {
        return x * x + y * y;
    }

    LL x, y, id;
}a[N];

int main(){
    scanf("%d", &n);
    for(int i = 1; i <= n; i++) {
        scanf("%lld%lld", &a[i].x, &a[i].y);
        id[i] = i;
    }

    while(1) {
        random_shuffle(id + 1, id + n);
        Vector now;
        for(int i = n; i >= 1; i--) {
            int pos = id[i];
            Vector nx1 = now + a[pos];
            Vector nx2 = now - a[pos];
            if(nx1.len() <= nx2.len()) {
                now = nx1;
                ans[pos] = 1;
            } else {
                now = nx2;
                ans[pos] = -1;
            }
        }

        if(now.len() <= base) {
            for(int i = 1; i <= n; i++) {
                printf("%d ", ans[i]);
            }
            puts("");
            return 0;
        }
    }
    return 0;
}

/*
*/