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DFS poj2488 A Knight's Journey

2015-07-20 17:23 507 查看
很经典的搜索+最小字典序路径打印

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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