poj-2488

//376K	16MS	G++
#include <stdio.h>
#include <string.h>
#define MAX 28

int board[MAX][MAX];
int p;
int q;
struct Point{
int x;
int y;
};

typedef struct Point Point;

Point visitOrder[MAX];

int visitedSquareNum;
int flag[MAX][MAX];
int squareNum = 0;

#define INF 999999;

int dfs(int curX, int curY, int visitedSquareNum);

int checkSquare(int curX, int curY, int visitedSquareNum) {
if (!flag[curX][curY]) {
int res = dfs(curX, curY, visitedSquareNum + 1);
return res;
}
return 0;
}

int dfs(int curX, int curY, int visitedSquareNum) {
// printf("dfs %d %d\n", curX, curY);
int maxVisted = 0;
flag[curX][curY] = 1;
visitOrder[visitedSquareNum].x = curX;
visitOrder[visitedSquareNum].y = curY;

if (visitedSquareNum == squareNum) {
return visitedSquareNum;
}

// x-1, y-2
if (curX - 1 >= 0 && curY - 2 >= 0) {
int res = checkSquare(curX - 1, curY - 2, visitedSquareNum);
if (res == squareNum) {
return squareNum;
}
}

//x+1, y-2
if (curX + 1 <= p - 1 && curY - 2 >= 0) {
int res = checkSquare(curX + 1, curY - 2, visitedSquareNum);
if (res == squareNum) {
return squareNum;
}
}

// x-2, y-1
if (curX - 2 >= 0 && curY - 1 >= 0) {
int res = checkSquare(curX - 2, curY - 1, visitedSquareNum);
if (res == squareNum) {
return squareNum;
}
}

//x+2, y-1
if (curX + 2 <= p - 1 && curY - 1 >= 0) {
int res = checkSquare(curX + 2, curY - 1, visitedSquareNum);
if (res == squareNum) {
return squareNum;
}
}

//x-2, y+1
if (curX - 2 >= 0 && curY + 1 <= q-1) {
int res = checkSquare(curX - 2, curY + 1, visitedSquareNum);
if (res == squareNum) {
return squareNum;
}
}

//x+2, y+1
if (curX + 2 <= p-1 && curY + 1 <= q-1) {
int res = checkSquare(curX + 2, curY + 1, visitedSquareNum);
if (res == squareNum) {
return squareNum;
}
}

//x-1, y+2
if (curX - 1 >= 0 && curY + 2 <= q-1) {
int res = checkSquare(curX - 1, curY + 2, visitedSquareNum);
if (res == squareNum) {
return squareNum;
}
}

//x+1, y+2
if (curX + 1 <= p-1 && curY + 2 <= q-1) {
int res = checkSquare(curX + 1, curY + 2, visitedSquareNum);
if (res == squareNum) {
return squareNum;
}
}

flag[curX][curY] = 0;
return 0; // no find a path to pass all
}

int travel(int curX, int curY) {
squareNum = p*q;
return dfs(curX, curY, 1);
}

void printPath() {
for (int i = 1; i <= squareNum; i++) {
printf("%c%d", visitOrder[i].y + 'A', visitOrder[i].x + 1);
}
printf("\n\n");
}

char findPath() {
int checkMaxX = p/2 + 1;
int checkMaxY = q/2 + 1;
char possible = 0;
for (int i = 0; i < checkMaxX; i++) { // only need check 1/4 board
for (int j = 0; j < checkMaxY; j++) {
memset(flag, 0, sizeof(flag));
if (travel(i, j)) { // if find a road, break;
// printf("%d %d\n",i, j);
return 1;
}
}
}
return 0;
}

void solve(int caseId) {
printf("Scenario #%d:\n", caseId);
if (findPath()) {
// printf("possible\n");
// printPath("");
printPath();
} else {
printf("impossible\n\n");
}
}

int main() {
int caseNum;
scanf("%d", &caseNum);
for (int i = 1; i <= caseNum; i++) {
scanf("%d %d", &p, &q);
solve(i);
}
}

流的题太飘忽了，先放放，刷刷简单的题。

题目本身DFS不难，坑的是字典序输出，不少好汉ＷＡ.

每次DFS有8个可选项，这8个选择要按照顺序进行判定，才能满足最后输出路径是字典序的要求。

每次DFS的8个判断都判断是否已经遍历完了所有的square，如果是，那么就直接return square的数量，否则return 0.

每次DFS除了传进去本次DFS开始的位置意外，还要传进去已经遍历的square的数量，这样才能在最后如果遍历完全部square的时候做判断是否完成。

同时搞一个数组来保存DFS的路径。最后输出做一下处理，将数字转化成ABCD。。即可.

这道题其实是验证了那个 国际象棋中马可以不重复跳完所有格子的 结论.