uva 10401 Injured Queen Problem(dp)
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题目链接:10401 - Injured Queen Problem
题目大意:给出一个字符串,要求在n * n(n为字符串的长度)的棋盘上摆放n个受伤的皇后,受伤的皇后只能攻击到同一列和它周围8个格子,如果字符串中第i个字符为'?'表示第i + 1个皇后可以摆放在任意行,如果为1 ~ F表示第i+1个皇后必须摆放在第str[i]行, 问,有多少种不同的摆法?
解题思路:一开始用递归 + 记忆化, 结果超时了, 后来发现其实可以写成递推,dp[i][j]代表第i个皇后摆放在第j行的摆法种类, dp[i][j] = dp[i - 1][k] ( 0 < k < n && abs (j - k) > 1)。
#include <stdio.h>#include <string.h>const int N = 20;char str[N];long long n, dp[N][N];int change(char c) {if (c >= '1' && c <= '9') return c - '1';else return c - 'A' + 9;;}void init() {n = strlen(str);memset(dp, 0, sizeof(dp));if (str[0] == '?') {for (int i = 0; i < n; i++)dp[0][i] = 1;}elsedp[0][change(str[0])] = 1;}long long solve() {init();for (int i = 1; i < n; i++) {if (str[i] == '?') {for (int j = 0; j < n; j++) {for (int k = 0; k < j - 1; k++)dp[i][j] += dp[i - 1][k];for (int k = j + 2; k < n; k++)dp[i][j] += dp[i - 1][k];}}else {int j = change(str[i]);for (int k = 0; k < j - 1; k++) dp[i][j] += dp[i - 1][k];for (int k = j + 2; k < n; k++)dp[i][j] += dp[i - 1][k];}}long long sum = 0;for (int i = 0; i < n; i++)sum += dp[n - 1][i];return sum;}int main() {while (scanf("%s", str) == 1) {printf("%lld\n", solve());}return 0;}
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