zoj 3823 Excavator Contest(构造)

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题目链接：zoj 3823 Excavator Contest

题目大意：一个人开着挖掘机要在N*N的格子上面移动，要求走完所有的格子，并且转完次数要至少为n*(n-1) - 1次，

并且终点和起点必须都在边界上。

解题思路：构造，因为终点和起点必须在边界上，进去的同时得留出一条路径出来。

奇数
偶数

奇数的情况分两种（图上两点所代表的正方形构造方式是一样的，即13，9为一类，11，7为一类）

偶数分为四类(图上两点所代表的正方形构造方式是不一样的，即14，12，10，8各为一类)

这是我做过最恶心的构造题了。

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 550;const int dir[4][2] = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};//const int G[5][50] = {{0}, {0}, {3, 4, 2, 1}, {5, 6, 9, 4, 7, 8, 3, 2, 1}};const int dir_down[4][4][4] = { {{3, 1, 2, 1}, {3, 3, 0, 3}, {1, 3, 0, 3}, {1, 3, 0, 0}},    {{2, 1, 3, 1}, {2, 1, 3, 3}, {0, 3, 1, 3}, {3, 0, 2, 0}},     {{3, 1, 2, 1}, {1, 3, 0, 3}, {1, 3, 0, 0}} };const int dir_left[5][4] = { {1, 2, 0, 2}, {1, 2, 0, 2}, {2, 1, 3, 1}, {0, 2, 1, 2}, {0, 2, 1, 1}};const int dir_up[5][4] = { {2, 0, 3, 0}, {2, 0, 3, 0}, {0, 2, 1, 2}, {3, 0, 2, 0}, {3, 0, 2, 2} };int L, R, g[maxn][maxn];void put(int n);inline void jump(int n, int& x, int& y, int& mv, int len, const int d[4]) {    for (int i = 0; i < n; i++) {        for (int j = 0; j < 4; j++) {            g[x][y] = mv;            mv += len;            x += dir[d[j]][0];            y += dir[d[j]][1];        }    }}inline void moveup(int n, int& x, int& y, int& mv, int len) {    //printf("moveup:%d\n", len);    if (n&1) {        jump(n / 2 - 1, x, y, mv, len, dir_up[0]);        for (int t = 0; t < 2; t++) {            g[x][y] = mv;            x += dir[2][0];            y += dir[2][1];            mv += len;        }    } else if ((n/2) % 4) {        jump(n / 2 - 2, x, y, mv, len, dir_up[1]);        jump(1, x, y, mv, len, dir_up[2]);    } else {        jump(n / 2 - 2, x, y, mv, len, dir_up[3]);        jump(1, x, y, mv, len, dir_up[4]);    }}inline void moveleft(int n, int& x, int& y, int& mv, int len) {    //printf("moveleft:%d\n", len);    if (n&1) {        jump(n / 2, x, y, mv, len, dir_left[0]);        for (int t = 0; t < 2; t++) { // down twice;            g[x][y] = mv;            x += dir[1][0];            y += dir[1][1];            mv += len;        }    } else if ((n/2) % 4) {        jump(n / 2 - 1, x, y, mv, len, dir_left[1]);        jump(1, x, y, mv, len, dir_left[2]);    } else {        jump(n / 2 - 1, x, y, mv, len, dir_left[3]);        jump(1, x, y, mv, len, dir_left[4]);    }}inline void movedown(int n, int& x, int& y, int& mv, int len) {    int p;    //printf("movedown!\n");    if (n&1) {        for (int k = 0; k < 4; k++) {            if (k == 0) p = n / 2;            else if (k == 1 || k == 3) p = 1;            else p = n / 2 - 2;            jump(p, x, y, mv, len, dir_down[0][k]);        }    } else if ((n/2) % 4 == 1) {        for (int k = 0; k < 4; k++) {            if (k == 0) p = n / 2 - 1;            else if (k == 1 || k == 3) p = (n == 2 && k == 3 ? 0 : 1);            else p = max(n / 2 - 2, 0);            jump(p, x, y, mv, len, dir_down[1][k]);        }    } else {        for (int k = 0; k < 3; k++) {            if (k == 0 || k == 1) p = n / 2 - 1;            else p = 1;            jump(p, x, y, mv, len, dir_down[2][k]);        }    }}void solve (int n, int sx, int sy, int ex, int ey, int flag) {    if (n <= 1) {        if (n == 1)            g[sx][sy] = L;        return;    }     /*    printf("%d:\n", n);    put(10);    */    if (n&1) {        if ((n/2)&1) {            if (flag) {                moveup(n, sx, sy, L, 1);                moveleft(n, ex, ey, R, -1);            } else {                moveup(n, ex, ey, R, -1);                moveleft(n, sx, sy, L, 1);            }            solve(n - 2, sx, sy, ex, ey, flag);        } else {            if (flag)                movedown(n, ex, ey, R, -1);            else                 movedown(n, sx, sy, L, 1);            solve(n - 2, sx, sy, ex, ey, flag^1);        }    } else {        if ((n/2)&1) {            if (flag)                movedown(n, ex, ey, R, -1);            else                movedown(n, sx, sy, L, 1);            solve(n - 2, sx, sy, ex, ey, flag^1);        } else {            if (flag) {                moveup(n, sx, sy, L, 1);                moveleft(n, ex, ey, R, -1);            } else {                moveup(n, ex, ey, R, -1);                moveleft(n, sx, sy, L, 1);            }            solve(n - 2, sx, sy, ex, ey, flag);        }    }}void put (int n) {    for (int i = 1; i <= n; i++) {        printf("%d", g[i][1]);        for (int j = 2; j <= n; j++)            printf(" %d", g[i][j]);        printf("\n");    }}int main () {    int cas, n;    scanf("%d", &cas);    while (cas--) {        scanf("%d", &n);        int sx, sy, ex, ey, flag;        L = 1, R = n * n;        if (n&1) {            if ((n/2)&1)                ex = 1, ey = sx = sy = n, flag = 1;            else                sx = sy = ex = 1, ey = n, flag = 0;        } else {            int t = n / 2;            t = (t - 1) % 4 + 1;            if (t == 1)                sx = 1, sy = ex = 2, ey = n, flag = 0;            else if (t == 2)                sx = sy = ey = n, ex = 1, flag = 1;            else if (t == 3)                sx = sy = ex = 1, ey = n, flag = 0;            else                sx = ey = n, sy = n - 1, ex = 2, flag = 1;        }        //printf("%d %d %d %d %d!!!\n", sx, sy, ex, ey, flag);        solve(n, sx, sy, ex, ey, flag);        put(n);    }    return 0;}

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