题目链接如下：
https://nanti.jisuanke.com/t/43321
思路：
显然我们要采用二分的方法来寻找答案，给定一个高度如果能确定在这个高度时是否可以安全到达终点，那我们就可以很快二分出最大可行的高度。在判断一个高度是否可行时，搜索从起点开始，在限制的高度下所有可以到达的坐标位置，检验是否经过终点即可。
/************************************************ ┆ ┏┓ ┏┓ ┆ ┆┏┛┻━━━┛┻┓ ┆ ┆┃ ┃ ┆ ┆┃ ━ ┃ ┆ ┆┃ ┳┛ ┗┳ ┃ ┆ ┆┃ ┃ ┆ ┆┃ ┻ ┃ ┆ ┆┗━┓ ┏━┛ ┆ ┆ ┃ ┃ ┆ ┆ ┃ ┗━━━┓ ┆ ┆ ┃ AC代马 ┣┓┆ ┆ ┃ ┏┛┆ ┆ ┗┓┓┏━┳┓┏┛ ┆ ┆ ┃┫┫ ┃┫┫ ┆ ┆ ┗┻┛ ┗┻┛ ┆ ************************************************ */ #include <iostream> #include <algorithm> #include <set> #include <vector> #include <queue> #include <string> #include <sstream> #include <cstdio> #include <cstring> #include<cctype> #include <math.h> #include <cmath> #include <bits/stdc++.h> #define INF 0x3f3f3f3f #define lb long double #define LINF 0x3f3f3f3f3f3f3f3f #define ull unsigned long long #define random(x) (rand()%x) #define sgn(x) (fabs(x) < eps ? 0 : ((x) < 0 ? -1 : 1)) #define rep(i, a, b) for(int i=a;i<b;i++) #define req(j, a, b) for(int j=a;j<b;j++) #define mem(a, b) memset(a, b, sizeof(a)) #define PI 3.141592653589793 #define K 20 using namespace std; typedef long long ll; const int maxn = 1e5 + 5; const double eps =1e-14; const double pi = acos(-1); const ll mod = 1e9 + 7; const int hash_mod = 19260817; inline int read() { int X=0,w=0; char ch=0; while(!isdigit(ch)) {w|=ch=='-';ch=getchar();} while(isdigit(ch)) X=(X<<3)+(X<<1)+(ch^48),ch=getchar(); return w?-X:X; } inline double dbread() { double X=0,Y=1.0; int w=0; char ch=0; while(!isdigit(ch)) {w|=ch=='-';ch=getchar();} while(isdigit(ch)) X=X*10+(ch^48),ch=getchar(); ch=getchar();//读入小数点 while(isdigit(ch)) X+=(Y/=10)*(ch^48),ch=getchar(); return w?-X:X; } inline void write(int x) { if(x<0) putchar('-'),x=-x; if(x>9) write(x/10); putchar(x%10+'0'); } ll gcd(ll a,ll b)//辗转相除法（欧几里德算法）求最大公约数 { return b ? gcd(b,a%b) : a; } ll lcm(ll a,ll b) { return a*b/gcd(a,b);//最小公倍数 } ll q_pow(ll base, ll p) { ll result = 1; while(p > 0) { if(p&1) {//此处等价于if(power%2==1) result = result * base % 1000; } p >>= 1;//此处等价于power=power/2 base = (base * base) % 1000; } return result; } const int N = 250005; int fri[N]; void init() { for(int i = 0; i <= N; i++) fri[i] = i; } int find(int a) //查找根节点 { int tem1 = a,tem2; while(a != fri[a]) a = fri[a]; while(tem1 != a) { tem2 = fri[tem1]; fri[tem1] = a; tem1 = tem2; } return a; } void Union(int a,int b) { int fri_a = find(a); int fri_b = find(b); if(fri_a != fri_b) { fri[fri_a] = fri_b; } return ; } int eval(char c) { if(c==' ') return 0; if(c=='\'') return 1; return c - 'A' + 2; } double pointSegDist(double cx, double cy, double px1, double py1, double px2, double py2) { cx -= px1; px2 -= px1; cy -= py1; py2 -= py1; double len = sqrt(px2*px2 + py2*py2); double ux = px2 / (len * 1.0); double uy = py2/ (len * 1.0); uy = -uy; double nx = cx * ux - cy * uy; double ny = cx * uy + cy * ux; if(nx>=0 && nx <= len) return abs(ny); if(nx > len) nx -= len; return sqrt(nx*nx + ny*ny); } bool f(ll n) { for(ll i=2;i<=n;i++) { if(n % (i * i) == 0) return false; } return true; } bool judgeStr(string s) { int len = s.size(); int i,j; for(i=0,j=len-1;i<=len/2;i++,j--) { if(s[i]!=s[j]) return 0; } return 1; } ll f(ll n, ll d) { ll ans = 0; while(n > 0 && n % d == 0) { n /= d; ans++; } return ans; } ll power(ll a, ll b) { long long ans = 1; for (int i=0;i<b;i++) ans *= a; return ans; } const int maxnn = 1e5+5; int dx[4]={1, 0, -1, 0}; int dy[4]={0, 1, 0, -1}; int vis[505][505]; int w[505][505]; int r, c; int bfs(int h) { mem(vis, 0); vis[0][0] = 1; queue<int> q; //x,y,dis q.push(0); q.push(0); q.push(0); while(q.size() > 0) { int x = q.front(); q.pop(); int y = q.front(); q.pop(); int dis = q.front(); q.pop(); if(w[x][y] - dis >= h) { if(x == r-1 && y == c-1) { return true; } for(int i=0;i<4;i++) { int a = x + dx[i]; int b = y + dy[i]; if(a >= 0 && a < r && b >= 0 && b < c && !vis[a][b]) { vis[a][b] = 1; q.push(a); q.push(b); q.push(dis+1); } } } } return false; } int main() { int t; cin >> t; while(t--) { cin >> r >> c; for(int i=0;i<r;i++) { for(int j=0;j<c;j++) { cin >> w[i][j]; } } int left = 0; int right = INF; while(left < right) { int mid = (left + right + 1) / 2; if(bfs(mid)) { left = mid; } else { right = mid - 1; } } if(left > 0) cout << left << endl; else cout << "impossible" << endl; } return 0; }