UVA - 639 Don't Get Rooked

题目大意：类似于八皇后。每行每列只能放一个位置，若有墙隔着就互不影响，问最多能放几个。

解题思路：回溯，先读入地图，ok() 分上下左右四个方向判断当前位置是否可以放置，ans 保存最佳答案。tmp 是当前放了多少个。遍历地图，若当前位置可放就标记一下，个数 ++，递归后还原状态。

#include<iostream> #include<cstdio>#include<cmath>#include<string.h>#include<stdlib.h>#include<algorithm>using namespace std;char map[5][5];bool vis[5][5];int n, ans, tmp;bool ok(int r, int c) {         int up, down, lef, rig;    up = down = lef = rig = 1;    for (int i = 0; i < r; i++) {        if (vis[i][c]) up = 0;        else if (map[i][c] == 'X') up = 1;    }    for (int i = r+1; i < n; i++) {        if (vis[i][c]) down = 0;        else if (map[i][c] == 'X') down = 1;    }    for (int i = 0; i < c; i++) {        if (vis[r][i]) lef = 0;        else if (map[r][i] == 'X') lef = 1;    }    for (int i = c+1; i < n; i++) {        if (vis[r][i]) rig = 0;        else if (map[r][i] == 'X') rig = 1;    }    if (up && down && lef && rig && !vis[r][c] && map[r][c] != 'X') return true;    else return false;}void search(int r, int c) {    if (c == n) { r++; c = 0;}    if (r == n) {        if (tmp > ans) ans = tmp;        return;    }    for (int i = r; i < n; i++)        for (int j = c; j < n; j++) {            int t = tmp;            if (ok(i, j)) {                tmp++;                vis[i][j] = 1;            }                search(i,j+1);                tmp = t;                vis[i][j] = 0;    }}int main() {    while (scanf("%d", &n) != EOF && n) {        memset(vis, 0, sizeof(vis));        for (int i = 0; i < n; i++)            scanf("%s", map[i]);        ans = 0;        tmp = 0;        search(0, 0);        printf("%d\n", ans);    }return 0; }