POJ 2110 Mountain Walking（二分+BFS）

思路：二分高度差mid且我们枚举的是区间上下界up和low.(mid=up-low)，对于一个区间[low,up]我们用BFS找,看看能不能找到一条从左上角到右下角的路,且路中走过的所有节点的高度值x都属于区间[low,up].

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
const int maxn=105;
int map[maxn][maxn];
int vis[maxn][maxn];
int n;
int dr[]={-1,1,0,0};//上下左右
int dc[]={0,0,-1,1};
struct Node
{
int r,c;
Node(int r,int c):r(r),c(c){}
};
bool BFS(int low,int up)
{
if(map[1][1]<low || map[1][1]>up) return false;
queue<Node> Q;
Q.push(Node(1,1));
vis[1][1]=1;
while(!Q.empty())
{
Node node=Q.front();Q.pop();
int r=node.r,c=node.c;
for(int dir=0;dir<4;dir++)
{
int nr=r+dr[dir] , nc=c+dc[dir];
if(nr>=1&&nr<=n&&nc>=1&&nc<=n&&!vis[nr][nc])
{
vis[nr][nc]=1;
if(map[nr][nc]>=low && map[nr][nc]<=up)
{
Q.push(Node(nr,nc));
if(nr==n&&nc==n) return true;
}
}
}
}
return false;
}
bool check(int d)
{
for(int low=0;low+d<=110;low++)
{
memset(vis,0,sizeof(vis));
if(BFS(low,low+d)) return true;
}
return false;
}
int main()
{
scanf("%d",&n);
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
scanf("%d",&map[i][j]);
int L=0,R=110;
while(R>L)
{
int mid=(R+L)>>1;
if(check(mid)) R=mid;
else L=mid+1;
}
printf("%d\n",R);
return 0;
}

Description

Farmer John and Bessie the cow have embarked on one of those 'active' vacations. They spend entire days walking in the mountains and then, at the end of the day, they tire and return to their vacation cabin.

Since climbing requires a lot of energy and they are already tired, they wish to return to the cabin using a path that has the least difference between its highest and lowest elevations, no matter how long that path is. Help FJ find this easy-to-traverse path.

The map of the mountains is given by an N x N (2 <= N <= 100) matrix of integer elevations (0 <= any elevation <= 110) FJ and Bessie are currently at the upper left position (row 1, column 1) and the cabin is at the lower right (row N, column N). They can travel
right, left, toward the top, or toward the bottom of the grid. They can not travel on a diagonal.

Input

* Line 1: The single integer, N

* Lines 2..N+1: Each line contains N integers, each of which specifies a square's height. Line 2 contains the first (top) row of the grid; line 3 contains the second row, and so on. The first number on the line corresponds to the first (left) column of the
grid, and so on.

Output

* Line 1: An integer that is the minimal height difference on the optimal path.

Sample Input

5
1 1 3 6 8
1 2 2 5 5
4 4 0 3 3
8 0 2 3 4
4 3 0 2 1

Sample Output

2