TOJ-1316 The Happy Worm

最新推荐文章于 2017-03-30 15:39:00 发布

转载最新推荐文章于 2017-03-30 15:39:00 发布 · 158 阅读

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/shenchuguimo/p/6349367.html

本文探讨了在一个m*n的矩形场地中，考虑到k个石头的位置限制，快乐蠕虫在睡眠时可能采取的不同位置数量。蠕虫可以水平或垂直伸展，但长度至少为2个单位，并且不能穿过石头或超出场地边界。通过遍历所有可能的位置并检查蠕虫的伸展方向，算法计算出所有合法位置的数量。

The Happy Worm lives in an m*n rectangular field. There are k stones placed in certain locations of the field. (Each square of the field is either empty, or contains a stone.) Whenever the worm sleeps, it lies either horizontally or vertically, and stretches so that its length increases as much as possible. The worm will not go in a square with a stone or out of the field. The happy worm can not be shorter than 2 squares.

The question you are to answer is how many different positions this worm could be in while sleeping.

Input

The first line of the input file contains a single integer t (1 ≤ t ≤ 11), the number of test cases, followed by the input data for each test case. The first line of each test case contains three integers m, n, and k (1 ≤ m,n ≤ 10000, 0 ≤ k ≤ 200000). The input for this test case will be followed by k lines. Each line contains two integers which specify the row and column of a stone. No stone will be given twice.

Output

There should be one line per test case containing the number of positions the happy worm can be in.

Sample Input

Sample Output

#include <iostream>
#include <memory.h>
using namespace std;
bool f[10010][10010];
int main(){
    int t,i,j,m,n,k,r,c,s,e,ans;
    cin>>t;
    while(t--){
        cin>>m>>n>>k;
        memset(f,0,sizeof(f));
        ans = 0;
        for(i=0;i<k;i++){
            cin>>r>>c;
            f[r][c] = true; 
        }
        for(i=0;i<=m+1;i++){
            f[i][0] = true;
            f[i][n+1] = true;
        }    
        for(i=0;i<=n+1;i++){
            f[0][i] = true;
            f[m+1][i] = true;
        }    
        for(i=1;i<=m;i++){
            for(j=1;j<=n+1;){
                while(f[i][j]){
                    j++;
                }
                if(j>n) break;
                if(!f[i][j]){
                    s = j;//cout<<"s "<<i<<" "<<j<<endl;
                    j++;
                    e = s;    
                }
                while(!f[i][j]){
                    e = j;
                    j++;
                }
                //cout<<"e "<<i<<" "<<j-1<<endl;
                if(e-s+1>1)    ans++; //cout<<e-s+1<<endl;     
            }
            
        }
        for(j=1;j<=n;j++){
            for(i=1;i<=m;){
                while(f[i][j]){
                    i++;
                }
                if(i>m)    break;
                if(!f[i][j]){
                    s = i;//cout<<"s "<<i<<" "<<j<<endl;
                    i++;
                    e = s;
                }
                while(!f[i][j]){
                    e = i;
                    i++;
                }//cout<<"e "<<i-1<<" "<<j<<endl;
                if(e-s+1>1)    ans++;      //cout<<e-s+1<<endl;
            }
            
        }
        cout<<ans<<endl;
    }
    return 0;
}

Source: Tehran 2004 Iran Nationwide

转载于:https://www.cnblogs.com/shenchuguimo/p/6349367.html