DFS poj2488 A Knight's Journey

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巧妙的利用LAST结构体，来保存上一个节点的位置，最后再递归输出

#include<cstdio>#include<cmath>#include<cstring>#include<queue>#include<vector>#include<functional>#include<algorithm>using namespace std;typedef long long LL;typedef pair<int, int> PII;const int MX = 30;const int INF = 0x3f3f3f3f;int m, n;bool vis[MX][MX];int dist[][2] = {{ -1, -2}, {1, -2}, { -2, -1}, {2, -1}, { -2, 1}, {2, 1}, { -1, 2}, {1, 2}};struct LAST {    int x, y;} La[MX][MX], End;bool DFS(int x, int y, int cnt) {    if(cnt == m * n) {        End.x = x;        End.y = y;        return true;    }    vis[x][y] = 1;    for(int k = 0; k < 8; k++) {        int nx = x + dist[k][0], ny = y + dist[k][1];        if(nx < 1 || nx > m || ny < 1 || ny > n) continue;        if(vis[nx][ny]) continue;        La[nx][ny].x = x;        La[nx][ny].y = y;        if(DFS(nx, ny, cnt + 1)) {            vis[x][y] = 0;            return true;        }    }    vis[x][y] = 0;    return false;}void print(int x, int y, int cnt) {    if(cnt == m * n) {        return;    }    print(La[x][y].x, La[x][y].y, cnt + 1);    printf("%c%d", 'A' + y - 1, x);    if(!cnt) printf("\n");}int main() {    int T, ansk = 0;    scanf("%d", &T);    while(T--) {        memset(vis, false, sizeof(vis));        End.x = -1;        scanf("%d%d", &m, &n);        bool sign = false;        for(int j = 1; j <= n; j++) {            for(int i = 1; i <= m; i++) {                if(DFS(i, j, 1)) {                    sign = true;                    break;                }            }            if(sign) break;        }        printf("Scenario #%d:\n", ++ansk);        if(End.x != -1) {            print(End.x, End.y, 0);        } else {            printf("impossible\n");        }        printf("\n");    }    return 0;}

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