炫酷迷宫
https://ac.nowcoder.com/acm/contest/331/C
C++版本一
題解：
std
比如4*4，K=10的時候，很容易構造出：
....
xxx.
.xx.
....
于是得到了一種繞圈的想法。
但是在比如5*2，K=7中，繞圈法又遇到了阻礙，不如直走換邊法，即
.x...
...x.
將這兩種方法合并起來，即可以得到一種看起來還不錯的解法。
如果能達到和以上算法同等優秀，就可以AC。
#include <bits/stdc++.h>using namespace std;//這份標程很不好看，建議閱讀其他人的程序。const int mn = 1005;int n, m; int k;char s[mn][mn];int calc(int n, int m) {int ans = 0;ans = n / 2 * (m + 1);if (n % 2 == 1) ans += m;return ans; }int lastx, lasty;void genn() {int ans = -1;for (int i = 1; i <= n; i += 2) {if (i / 2 % 2 == 0) {for (int j = 1; j <= m; j++) {s[i][j] = '.';ans++;if (ans == k) {lastx = i, lasty = j;return;}}s[i + 1][m] = '.';ans++;if (ans == k) {lastx = i + 1, lasty = m;return;}} else {for (int j = m; j; j--) {s[i][j] = '.';ans++;if (ans == k) {lastx = i, lasty = j;return;}}s[i + 1][1] = '.';ans++;if (ans == k) {lastx = i + 1, lasty = 1;return;}}} }void genm() {int ans = -1;for (int i = 1; i <= m; i += 2) {if (i / 2 % 2 == 0) {for (int j = 1; j <= n; j++) {s[j][i] = '.';ans++;if (ans == k) {lastx = j, lasty = i;return;}}s[n][i + 1] = '.';ans++;if (ans == k) {lastx = n, lasty = i + 1;return;}} else {for (int j = n; j; j--) {s[j][i] = '.';ans++;if (ans == k) {lastx = j, lasty = i;return;}}s[1][i + 1] = '.';ans++;if (ans == k) {lastx = 1, lasty = i + 1;return;}}} }void app() {int u, d, l, r;u = 3, l = 1;d = n, r = m;int x = 1, y = 1;int c = 0;int step = 1;while (step) {if (c % 4 == 0) { // rightstep = max(min(k, r - y), 0);for (int i = y; i <= y + step; i++) s[x][i] = '.';y += step;r -= 2;} else if (c % 4 == 1) { // downstep = max(min(k, d - x), 0);for (int i = x; i <= x + step; i++) s[i][y] = '.';x += step;d -= 2;} else if (c % 4 == 2) { // leftstep = max(0, min(k, y - l));for (int i = y; i >= y - step; i--) s[x][i] = '.';y -= step;l += 2;} else if (c % 4 == 3) { // upstep = max(0, min(k, x - u));for (int i = x; i >= x - step; i--) s[i][y] = '.';x -= step;u += 2;}k -= step;c++;}lastx = x, lasty = y; }int main() {cin >> n >> m >> k;for (int i = 1; i <= n; i++) {for (int j = 1; j <= m; j++) s[i][j] = 'x';}if (max(calc(n, m), calc(m, n)) - 1 >= k) {if (calc(n, m) >= calc(m, n)) genn();else genm();}else app();printf("%d %d\n%d %d\n", 1, 1, lastx, lasty);for (int i = 1; i <= n; i++) puts(s[i] + 1);return 0; }C++版本二
#include <map> #include <set> #include <ctime> #include <cmath> #include <queue> #include <stack> #include <vector> #include <string> #include <cstdio> #include <cstdlib> #include <cstring> #include <sstream> #include <iostream> #include <algorithm> #include <functional> using namespace std; #define For(i, x, y) for(int i=x;i<=y;i++) #define _For(i, x, y) for(int i=x;i>=y;i--) #define Mem(f, x) memset(f,x,sizeof(f)) #define Sca(x) scanf("%d", &x) #define Sca2(x,y) scanf("%d%d",&x,&y) #define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define Scl(x) scanf("%lld",&x); #define Pri(x) printf("%d\n", x) #define Prl(x) printf("%lld\n",x); #define CLR(u) for(int i=0;i<=N;i++)u[i].clear(); #define LL long long #define ULL unsigned long long #define mp make_pair #define PII pair<int,int> #define PIL pair<int,long long> #define PLL pair<long long,long long> #define pb push_back #define fi first #define se second typedef vector<int> VI; int read(){int x = 0,f = 1;char c = getchar();while (c<'0' || c>'9'){if (c == '-') f = -1;c = getchar();}while (c >= '0'&&c <= '9'){x = x * 10 + c - '0';c = getchar();}return x*f; } const double eps = 1e-9; const int maxn = 1010; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; int N,M,K; const int a[4][2] = {0,1,0,-1,1,0,-1,0}; char MAP[maxn][maxn]; char MAP2[maxn][maxn]; bool vis[maxn][maxn]; struct node{int x,y,num;node(int x,int y,int num):x(x),y(y),num(num){} }; bool check(node a){return 1 <= a.x && a.x <= N && 1 <= a.y && a.y <= M && MAP[a.x][a.y] == '.' && !vis[a.x][a.y]; } void BFS(){queue<node>Q;Q.push(node(1,1,0));vis[1][1] = 1;while(!Q.empty()){node u = Q.front(); Q.pop();if(u.num == K){printf("%d %d\n",u.x,u.y);return;}for(int i = 0; i < 4; i ++){node h = node(u.x + a[i][0],u.y + a[i][1],u.num + 1);if(!check(h)) continue;Q.push(h);vis[h.x][h.y] = 1;}} } int main(){Sca3(N,M,K);int ans = 0;for(int i = 1; i <= N ; i ++){for(int j = 1; j <= M ; j ++){if(i & 1){MAP[i][j] = '.';ans++;}else MAP[i][j] = 'x';}}for(int i = 2; i <= N ;i += 4){MAP[i][M] = '.';ans++;}for(int i = 4; i <= N; i += 4){MAP[i][1] = '.';ans++;}for(int i = 1; i <= M ; i ++){for(int j = 1; j <= N ; j ++){if(i & 1){MAP2[j][i] = '.';ans--;}else MAP2[j][i] = 'x';}}for(int i = 2; i <= M ; i += 4){MAP2[N][i] = '.';ans--;}for(int i = 4; i <= M ; i += 4){MAP2[1][i] = '.';ans--;}if(ans < 0){For(i,1,N) For(j,1,M) MAP[i][j] = MAP2[i][j];}printf("1 1\n");BFS();for(int i = 1; i <= N ; i ++){for(int j = 1; j <= M ; j++){printf("%c",MAP[i][j]);}puts("");}return 0; }?
總結
