UVA 10401Injured Queen Problem （dp）

题意:

放n∗n棋盘放n个互不攻击Injured Queen有多少种方法

分析:

dp[i][j]:=第i列放置皇后在第j行的方法数
转移就很暴力了,枚举上一列放置的位置,判断下合法性,把dp[1][i]先跑出来

代码:

////  Created by TaoSama on 2015-11-04//  Copyright (c) 2015 TaoSama. All rights reserved.////#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;#define pr(x) cout << #x << " = " << x << "  "#define prln(x) cout << #x << " = " << x << endlconst int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;typedef long long LL;char s[20];LL n, dp[20][20];int get(char c) {    if(isdigit(c)) return c - '0';    else return c - 'A' + 10;}int main() {#ifdef LOCAL    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    while(scanf("%s", s + 1) == 1) {        n = strlen(s + 1);        memset(dp, 0, sizeof dp);        if(s[1] == '?') for(int i = 1; i <= n; ++i) dp[1][i] = 1;        else dp[1][get(s[1])] = 1;        for(int i = 2; i <= n; ++i) {            if(s[i] == '?') {                for(int j = 1; j <= n; ++j) {                    for(int k = 1; k <= n; ++k)                        if(k != j && k != j - 1 && k != j + 1)                            dp[i][j] += dp[i - 1][k];                }            } else {                int j = get(s[i]);                for(int k = 1; k <= n; ++k)                    if(k != j && k != j - 1 && k != j + 1)                        dp[i][j] += dp[i - 1][k];            }        }        LL ans = 0;        for(int i = 1; i <= n; ++i) ans += dp[n][i];        printf("%lld\n", ans);    }    return 0;}