POJ-2110 Mountain Walking

Mountain Walking

二分 + \(bfs\)

最终要求的最低差

二分答案，枚举最低海拔，这样就有一个能够接受的海拔范围 \([l, r]\)

根据这个范围，看看 \(bfs\) 能不能搜索到 \((n, n)\)

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
using namespace std;
const int maxn = 120;
#define pii pair<int, int>
int num[maxn][maxn], n, vis[maxn][maxn];
const int xi[4] = {0, 0, 1, -1};
const int yi[4] = {1, -1, 0, 0};

bool check(int x, int y)
{
    return x >= 1 && y >= 1 && x <= n && y <= n && vis[x][y] == 0;
}

bool bfs(int l, int r)
{
    queue<pii>q;
    for(int i=0; i<=n; i++)
        for(int j=0; j<=n; j++)
            vis[i][j] = 0;
    if(num[1][1] <= r && num[1][1] >= l) q.push(make_pair(1, 1));
    while(q.size())
    {
        pii now = q.front();
        q.pop();
        if(vis[now.first][now.second]) continue;
        vis[now.first][now.second] = 1;
        for(int i=0; i<4; i++)
        {
            int xx = now.first + xi[i];
            int yy = now.second + yi[i];
            if(check(xx, yy) && num[xx][yy] >= l && num[xx][yy] <= r)
                q.push(make_pair(xx, yy));
        }
    }
    return vis[n][n];
}

bool query(int len)
{
    for(int i=len; i<=110; i++)
        if(bfs(i - len, i)) 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", &num[i][j]);
    int l = 0, r = 110;
    while(l < r)
    {
        int mid = l + r >> 1;
        if(query(mid)) r = mid;
        else l = mid + 1;
    }
    printf("%d\n", l);
    return 0;
}