Land oj 1606 - Funny Sheep（技巧&规律）

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Problem 1606 - Funny Sheep

Time Limit: 1000MS Memory Limit: 65536KB
Total Submit: 612 Accepted: 169 Special Judge: No

Description

There are N+1 rows and M+1 columns fence with N*M grids on the grassland. Each grid has a sheep. In order to let the sheep together, we need to dismantle the fence. Every time you can remove a row or a column of fences. What’s the least number of times to reach the goal?

Input

There are multiple test cases.
The first line of each case contains two integers N and M. (1≤N,M≤1000)

Output

For each case, output the answer in a line.

Sample Input

1 2
2 2

Sample Output

1
2

//题意：

现在有一个n*m的矩形的网格栅栏，每个网格里有一头羊，问最少拆掉多少根的栅栏可以让所有的羊走到一起。

//思路：

因为是让所有的羊走到一起，所以通过模拟可以找出规律：拆的最少栅栏数为min(min(n,m),m+n-2);n表示将横行拆掉，只剩下一行

m表示将所有的纵列拆掉，只剩下一列，n+m-2表示将中间的栅栏拆掉。找到它们三个之间的最小值即可。

#include<stdio.h>#include<string.h>#include<math.h>#include<map>#include<queue>#include<stack>#include<set>#include<algorithm>#include<iostream>#define INF 0x3f3f3f3f#define ull unsigned long long#define ll long long#define IN __int64#define N 100010#define M 1000000007using namespace std;int main(){int n,m;while(scanf("%d%d",&n,&m)!=EOF){int ans=min(n,m);ans=min(ans,n+m-2);printf("%d\n",ans);}return 0;}

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