https://ac.nowcoder.com/acm/contest/318/F

C++版本一

題解：思路：二分+搜索，每次二分利姆魯?shù)某跏脊糁?#xff0c;如果這個攻擊值能夠拯救到靜就將mid賦值給ub，同時令ans = mid，否則就將mid賦值給lb，最后輸出ans即可。

#include<bits/stdc++.h> using namespace std;int t, n, sx, sy, ex, ey; int mp[505][505], vis[505][505]; int dx[] = {1, -1, 0, 0}, dy[] = {0, 0, 1, -1};struct node {int x, y; }nw, nxt;bool go(int x, int y) {if(x < 1 || x > n) return false;if(y < 1 || y > n) return false;if(vis[x][y]) return false;return true; }bool check(int x) {queue<node> q;memset(vis, 0, sizeof(vis));nw.x = sx, nw.y = sy;vis[sx][sy] = 1;q.push(nw);while(!q.empty()) {nw = q.front(); q.pop();if(nw.x == ex && nw.y == ey) return true;for(int i = 0; i < 4; i++) {nxt.x = nw.x + dx[i];nxt.y = nw.y + dy[i];if(go(nxt.x, nxt.y) && mp[nxt.x][nxt.y] <= x) {q.push(nxt);vis[nxt.x][nxt.y] = 1;}}}return false; }int main() {scanf("%d", &t);while(t--) {int mx = -1;scanf("%d", &n);for(int i = 1; i <= n; i++) {for(int j = 1; j <= n; j++) {scanf("%d", &mp[i][j]);mx = max(mx, mp[i][j]);}}scanf("%d%d%d%d", &sx, &sy, &ex, &ey);int ub = mx, lb = 0, mid;while(ub >= lb) {mid = (ub + lb) >> 1;if(check(mid)) {ub = mid - 1;} else {lb = mid + 1;}}printf("%d\n", lb);}return 0; }

C++版本二

題解：

二分+BFS

/* *@Author: STZG *@Language: C++ */ #include <bits/stdc++.h> #include<iostream> #include<algorithm> #include<cstdlib> #include<cstring> #include<cstdio> #include<string> #include<vector> #include<bitset> #include<queue> #include<deque> #include<stack> #include<cmath> #include<list> #include<map> #include<set> //#define DEBUG #define RI register int using namespace std; typedef long long ll; //typedef __int128 lll; const int N=1000+10; const int MOD=1e9+7; const double PI = acos(-1.0); const double EXP = 1E-8; const int INF = 0x3f3f3f3f; int t,n,m,k; int a[N][N]; int b[][2]={1,0,0,1,-1,0,0,-1}; int vis[N][N]; int sx,sy,ex,ey; struct node{int x,y; }f,s,tmp; bool bfs(int x){memset(vis,0,sizeof(vis));queue<node>q;s.x=sx;s.y=sy;vis[sx][sy]=1;q.push(s);while(!q.empty()){f=q.front();q.pop();if(f.x==ex&&f.y==ey){return 1;}for(int i=0;i<4;i++){tmp.x=f.x+b[i][0];tmp.y=f.y+b[i][1];//cout<<tmp.x<<" "<<tmp.y<<endl;if(1<=tmp.x&&tmp.x<=n&&1<=tmp.y&&tmp.y<=n&&!vis[tmp.x][tmp.y]&&x>=a[tmp.x][tmp.y]){vis[tmp.x][tmp.y]=1;q.push(tmp);}}}return 0; } int main() { #ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout); #endifwhile(~scanf("%d",&t)){while(t--){scanf("%d",&n);for(int i=1;i<=n;i++){for(int j=1;j<=n;j++){scanf("%d",&a[i][j]);}}scanf("%d%d%d%d",&sx,&sy,&ex,&ey);int l=0;int r=1e5;int mid;int ans=0;while(l<=r){mid=(l+r)>>1;if(bfs(mid)){ans=mid;r=mid-1;}else{l=mid+1;}}printf("%d\n",ans);}}//cout << "Hello world!" << endl;return 0; }

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