https://ac.nowcoder.com/acm/contest/331/C

C++版本一

题解：

std

比如4*4，K=10的时候，很容易构造出：

....

xxx.

.xx.

....

于是得到了一种绕圈的想法。

但是在比如5*2，K=7中，绕圈法又遇到了阻碍，不如直走换边法，即

.x...

...x.

将这两种方法合并起来，即可以得到一种看起来还不错的解法。

如果能达到和以上算法同等优秀，就可以AC。

#include <bits/stdc++.h>
 
using namespace std;
 
//这份标程很不好看，建议阅读其他人的程序。
 
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) {  // right
            step = 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) {  // down
            step = 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) {  // left
            step = 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) {  // up
            step = 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;
}