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  • [AHOI2009]中国象棋 BZOJ1801 dp

    题目描述

    这次小可可想解决的难题和中国象棋有关,在一个N行M列的棋盘上,让你放若干个炮(可以是0个),使得没有一个炮可以攻击到另一个炮,请问有多少种放置方法。大家肯定很清楚,在中国象棋中炮的行走方式是:一个炮攻击到另一个炮,当且仅当它们在同一行或同一列中,且它们之间恰好 有一个棋子。你也来和小可可一起锻炼一下思维吧!

    输入输出格式

    输入格式:

    一行包含两个整数N,M,之间由一个空格隔开。

    输出格式:

    总共的方案数,由于该值可能很大,只需给出方案数模9999973的结果。

    输入输出样例

    输入样例#1: 复制
    1 3
    输出样例#1: 复制
    7

    说明

    样例说明

    除了3个格子里都塞满了炮以外,其它方案都是可行的,所以一共有2*2*2-1=7种方案。

    数据范围

    100%的数据中N和M均不超过100

    50%的数据中N和M至少有一个数不超过8

    30%的数据中N和M均不超过6

    考验思维的一道dp;

    #include<iostream>
    #include<cstdio>
    #include<algorithm>
    #include<cstdlib>
    #include<cstring>
    #include<string>
    #include<cmath>
    #include<map>
    #include<set>
    #include<vector>
    #include<queue>
    #include<bitset>
    #include<ctime>
    #include<time.h>
    #include<deque>
    #include<stack>
    #include<functional>
    #include<sstream>
    //#include<cctype>
    //#pragma GCC optimize(2)
    using namespace std;
    #define maxn 200005
    #define inf 0x7fffffff
    //#define INF 1e18
    #define rdint(x) scanf("%d",&x)
    #define rdllt(x) scanf("%lld",&x)
    #define rdult(x) scanf("%lu",&x)
    #define rdlf(x) scanf("%lf",&x)
    #define rdstr(x) scanf("%s",x)
    #define mclr(x,a) memset((x),a,sizeof(x))
    typedef long long  ll;
    typedef unsigned long long ull;
    typedef unsigned int U;
    #define ms(x) memset((x),0,sizeof(x))
    const long long int mod = 9999973;
    #define Mod 1000000000
    #define sq(x) (x)*(x)
    #define eps 1e-5
    typedef pair<int, int> pii;
    #define pi acos(-1.0)
    //const int N = 1005;
    #define REP(i,n) for(int i=0;i<(n);i++)
    typedef pair<int, int> pii;
    
    inline int rd() {
    	int x = 0;
    	char c = getchar();
    	bool f = false;
    	while (!isdigit(c)) {
    		if (c == '-') f = true;
    		c = getchar();
    	}
    	while (isdigit(c)) {
    		x = (x << 1) + (x << 3) + (c ^ 48);
    		c = getchar();
    	}
    	return f ? -x : x;
    }
    
    
    ll gcd(ll a, ll b) {
    	return b == 0 ? a : gcd(b, a%b);
    }
    int sqr(int x) { return x * x; }
    
    
    
    /*ll ans;
    ll exgcd(ll a, ll b, ll &x, ll &y) {
    	if (!b) {
    		x = 1; y = 0; return a;
    	}
    	ans = exgcd(b, a%b, x, y);
    	ll t = x; x = y; y = t - a / b * y;
    	return ans;
    }
    */
    
    int n, m, ans;
    ll dp[200][200][200];
    int C(int x) {
    	return (x*(x - 1) / 2) % mod;
    }
    int main()
    {
    //	ios::sync_with_stdio(0);
    	n = rd(); m = rd();
    	dp[0][0][0] = 1;
    	for (int i = 1; i <= n; i++) {
    		for (int j = 0; j <= m; j++) {
    			for (int k = 0; k + j <= m; k++) {
    				dp[i][j][k] = dp[i - 1][j][k];
    				if (k >= 1) {
    					dp[i][j][k] =1ll* (dp[i][j][k] + dp[i - 1][j + 1][k - 1] * (j + 1));
    				}
    				if (j >= 1) {
    					dp[i][j][k] = 1ll*(dp[i][j][k] + dp[i - 1][j - 1][k] * (m - (j - 1) - k));
    				}
    				if (k >= 2) {
    					dp[i][j][k] = 1ll*(dp[i][j][k] + dp[i - 1][j + 2][k - 2] * ((j + 2)*(j + 1) / 2));
    				}
    				if (k >= 1) {
    					dp[i][j][k] = 1ll*(dp[i][j][k] + dp[i - 1][j][k - 1] * j*(m - (k - 1) - j));
    				}
    				if (j >= 2) {
    					dp[i][j][k] = 1ll*(dp[i][j][k] + dp[i - 1][j - 2][k] * C(m - (j - 2) - k));
    				}
    				dp[i][j][k] %= 1ll*mod;
    			}
    		}
    	}
    //	int ans = 0;
    	for (int i = 0; i <= m; i++) {
    		for (int j = 0; j <= m; j++) {
    			ans = 1ll*(ans + dp[n][i][j]) % mod;
    		}
    	}
    	printf("%lld
    ", 1ll * ans%mod);
    	return 0;
    }
    
    EPFL - Fighting
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  • 原文地址:https://www.cnblogs.com/zxyqzy/p/10354593.html
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