POJ 1321 棋盘问题（暴力dfs or DP）

来源：互联网发布：淘宝商家登陆编辑：程序博客网时间：2024/06/03 14:47

POJ1321 棋盘问题

题目链接

题意：中文题不多说

思路：本来是开二分图匹配的专题的，结果开错题开了这题- -，然后看了下数据范围，暴力爆过，额，不过没跑到0MS，于是又用状压DP搞了一发搞到0MS了

代码：

暴力：

#include <cstdio>#include <cstring>#include <vector>using namespace std;const int N = 10;int n, k, vis[N];char str[N];vector<int> g[N];int dfs(int u, int sum) {if (sum == k) return 1;if (u == n) return 0;int ans = 0;for (int i = 0; i < g[u].size(); i++) {if (vis[g[u][i]]) continue;vis[g[u][i]] = 1;ans += dfs(u + 1, sum + 1);vis[g[u][i]] = 0;}ans += dfs(u + 1, sum);return ans;}int main() {while (~scanf("%d%d", &n, &k)) {if (n == -1 && k == -1) break;for (int i = 0; i < n; i++) {g[i].clear();scanf("%s", str);for (int j = 0; j < n; j++) {if (str[j] == '#')g[i].push_back(j);}}printf("%d\n", dfs(0, 0));}return 0;}

DP：

#include <cstdio>#include <cstring>#include <vector>using namespace std;const int N = 10;int n, K;char str[N];vector<int> g[N];int dp[N][1<<N];int bitcount(int x) {return x == 0 ? 0 : bitcount(x>>1) + (x&1);}int main() {while (~scanf("%d%d", &n, &K) && n != -1) {for (int i = 1; i <= n; i++) {g[i].clear();scanf("%s", str);for (int j = 0; j < n; j++) {if (str[j] == '#')g[i].push_back(j);}}dp[0][0] = 1;int ans = 0;for (int i = 1; i <= n; i++) {for (int j = 0; j < (1<<n); j++) {dp[i][j] = dp[i - 1][j];for (int k = 0; k < g[i].size(); k++) {if (j&(1<<g[i][k])) {dp[i][j] += dp[i - 1][j^(1<<g[i][k])];}}if (i == n && bitcount(j) == K)ans += dp[i][j];}}printf("%d\n", ans);}return 0;}

0 0