CodeForces 350C Bombs(模拟)

最新推荐文章于 2021-09-18 12:18:43 发布

77458

最新推荐文章于 2021-09-18 12:18:43 发布

阅读量949

点赞数 1

分类专栏： ACM_模拟题

本文链接：https://blog.youkuaiyun.com/qq_18661257/article/details/51913741

版权

ACM_模拟题专栏收录该内容

54 篇文章

订阅专栏

C - Bombs

Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 350C

Description

You've got a robot, its task is destroying bombs on a square plane. Specifically, the square plane contains n bombs, the i-th bomb is at point with coordinates (x_i, y_i). We know that no two bombs are at the same point and that no bomb is at point with coordinates (0, 0). Initially, the robot is at point with coordinates (0, 0). Also, let's mark the robot's current position as (x, y). In order to destroy all the bombs, the robot can perform three types of operations:

Operation has format "1 k dir". To perform the operation robot have to move in direction dirk (k ≥ 1) times. There are only 4directions the robot can move in: "R", "L", "U", "D". During one move the robot can move from the current point to one of following points: (x + 1, y), (x - 1, y), (x, y + 1), (x, y - 1) (corresponding to directions). It is forbidden to move from point (x, y), if at least one point on the path (besides the destination point) contains a bomb.
Operation has format "2". To perform the operation robot have to pick a bomb at point (x, y) and put it in a special container. Thus, the robot can carry the bomb from any point to any other point. The operation cannot be performed if point (x, y) has no bomb. It is forbidden to pick a bomb if the robot already has a bomb in its container.
Operation has format "3". To perform the operation robot have to take a bomb out of the container and destroy it. You are allowed to perform this operation only if the robot is at point (0, 0). It is forbidden to perform the operation if the container has no bomb.

Help the robot and find the shortest possible sequence of operations he can perform to destroy all bombs on the coordinate plane.

Input

The first line contains a single integer n (1 ≤ n ≤ 10⁵) — the number of bombs on the coordinate plane. Next n lines contain two integers each. The i-th line contains numbers (x_i, y_i) ( - 10⁹ ≤ x_i, y_i ≤ 10⁹) — the coordinates of the i-th bomb. It is guaranteed that no two bombs are located at the same point and no bomb is at point (0, 0).

Output

In a single line print a single integer k — the minimum number of operations needed to destroy all bombs. On the next lines print the descriptions of these k operations. If there are multiple sequences, you can print any of them. It is guaranteed that there is the solution where k ≤ 10⁶.

Sample Input

Input

2
1 1
-1 -1

Output

Input

Output

题意:给你一个机器人，在(0,0),在除了(0,0)点之外的点中存在一些炸弹，你的任务是将这些炸弹移到(0,0)处销毁

直接用一个数组或者vector,将所有的炸弹加入其中，然后排序，接着一个个的销毁，如此几个步骤是不会变的:

1.到达炸弹位置

2.将炸弹移动到(0,0)

3.销毁炸弹

最终输出最少要多少步?

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <vector>

using namespace std;
const int MAXN = 1e5 + 5;
struct o{
    int x, y;
    o(int x,int y):x(x), y(y){}
    bool operator < (const o &a) const{
        return abs(x) + abs(y) < abs(a.x) + abs(a.y);//绝对值排序
    }
};
vector<o> V;
int A[MAXN], n, x, y;
int main(){
    while(~scanf("%d", &n)){
        int ans = 0;
        V.clear();
        for(int i = 0;i < n;i ++){
            scanf("%d%d", &x, &y);
            if(x != 0) ans ++;//判断这个点是一步到还是两步到
            if(y != 0) ans ++;
            V.push_back(o(x, y));
        }
        ans *= 2;//一去一回
        ans += n * 2;//2操作和3操作各一回
        sort(V.begin(), V.end());
        printf("%d\n", ans);
        for(int i = 0;i < n;i ++){
            if(V[i].y > 0) printf("1 %d U\n", abs(V[i].y));
            if(V[i].y < 0) printf("1 %d D\n", abs(V[i].y));
            if(V[i].x > 0) printf("1 %d R\n", abs(V[i].x));
            if(V[i].x < 0) printf("1 %d L\n", abs(V[i].x));
            printf("2\n");
            if(V[i].y > 0) printf("1 %d D\n", abs(V[i].y));
            if(V[i].y < 0) printf("1 %d U\n", abs(V[i].y));
            if(V[i].x > 0) printf("1 %d L\n", abs(V[i].x));
            if(V[i].x < 0) printf("1 %d R\n", abs(V[i].x));
            printf("3\n");
        }
    }
    return 0;
}