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  • Codeforces Round #335 (Div. 2) B

    B. Testing Robots
    time limit per test
    2 seconds
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    The Cybernetics Failures (CF) organisation made a prototype of a bomb technician robot. To find the possible problems it was decided to carry out a series of tests. At the beginning of each test the robot prototype will be placed in cell (x0, y0) of a rectangular squared field of size x × y, after that a mine will be installed into one of the squares of the field. It is supposed to conduct exactly x·y tests, each time a mine is installed into a square that has never been used before. The starting cell of the robot always remains the same.

    After placing the objects on the field the robot will have to run a sequence of commands given by string s, consisting only of characters 'L', 'R', 'U', 'D'. These commands tell the robot to move one square to the left, to the right, up or down, or stay idle if moving in the given direction is impossible. As soon as the robot fulfills all the sequence of commands, it will blow up due to a bug in the code. But if at some moment of time the robot is at the same square with the mine, it will also blow up, but not due to a bug in the code.

    Moving to the left decreases coordinate y, and moving to the right increases it. Similarly, moving up decreases the x coordinate, and moving down increases it.

    The tests can go on for very long, so your task is to predict their results. For each k from 0 to length(s) your task is to find in how many tests the robot will run exactly k commands before it blows up.

    Input

    The first line of the input contains four integers xyx0, y0 (1 ≤ x, y ≤ 500, 1 ≤ x0 ≤ x, 1 ≤ y0 ≤ y) — the sizes of the field and the starting coordinates of the robot. The coordinate axis X is directed downwards and axis Y is directed to the right.

    The second line contains a sequence of commands s, which should be fulfilled by the robot. It has length from 1 to 100 000 characters and only consists of characters 'L', 'R', 'U', 'D'.

    Output

    Print the sequence consisting of (length(s) + 1) numbers. On the k-th position, starting with zero, print the number of tests where the robot will run exactly k commands before it blows up.

    Sample test(s)
    input
    3 4 2 2
    UURDRDRL
    output
    1 1 0 1 1 1 1 0 6
    input
    2 2 2 2
    ULD
    output
    1 1 1 1
    Note

    In the first sample, if we exclude the probable impact of the mines, the robot's route will look like that: .

    这题的意思是没有走过的,如果走到了,就算一次 。以前走过的,现在再走一次,不会爆炸。始终没有走过的,一共有多少种-已经走过的,就是可能爆炸的可能次数

    #include<stdio.h>
    //#include<bits/stdc++.h>
    #include<string.h>
    #include<iostream>
    #include<math.h>
    #include<sstream>
    #include<set>
    #include<queue>
    #include<map>
    #include<vector>
    #include<algorithm>
    #include<limits.h>
    #define inf 0x3fffffff
    #define INF 0x3f3f3f3f
    #define lson l,m,rt<<1
    #define rson m+1,r,rt<<1|1
    #define LL long long
    #define ULL unsigned long long
    using namespace std;
    int i,j;
    int n,m;
    int sum,ans,flag;
    int a[510][510];
    int x,y,x0,y0;
    string s;
    int main()
    {
        cin>>x>>y>>x0>>y0;
        cin>>s;
        sum=0;
        for(i=0; i<s.length(); i++)
        {
            if(!a[x0][y0])
            {
                ++sum;
                cout<<"1"<<" ";
                a[x0][y0]=1;
            }
            else
            {
                cout<<"0"<<" ";
            }
            if(s[i]=='L')
            {
                y0--;
            }
            else if(s[i]=='R')
            {
                y0++;
            }
            else if(s[i]=='U')
            {
                x0--;
            }
            else if(s[i]=='D')
            {
                x0++;
            }
            if(x0<=0)
            {
                x0=1;
            }
            else if(x0>x)
            {
                x0=x;
            }
            if(y0<=0)
            {
                y0=1;
            }
            else if(y0>y)
            {
                y0=y;
            }
        }
        cout<<x*y-sum<<endl;
        return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/yinghualuowu/p/5046442.html
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