2017 ICPC区域赛（西安站）--- J题 LOL（DP）

problem description

5 friends play LOL together . Every one should BAN one character and PICK one character . The enemy should BAN 55 characters and PICK 55 characters . All these 2020 heroes must be different .

Every one can BAN any heroes by his personal washes . But he can only PICK heroes which he has bought .

Suppose the enemy can PICK or BAN any heroes. How many different ways are there satisfying the conditions?

For example , a valid way is :

Player 11 : picks hero 11, bans hero 22

Player 22 : picks hero 33, bans hero 44

Player 33 : picks hero 5, bans hero 66

Player 44 : picks hero 77, bans hero 88

Player 55 : picks hero 99, bans hero 1010

Enemies pick heroes 11,12,13,14,1511,12,13,14,15 , ban heroes 16,17,18,19,2016,17,18,19,20 .

Input

The input contains multiple test cases.(No more than 2020)

In each test case . there’s 55 strings S[1] \sim S[5]S[1]∼S[5] ,respectively whose lengths are 100100 , For the ii-th person if he has bought the jj-th hero, the jj-th character of S[i]S[i] is '11', or '00' if not. The total number of heroes is exactly 100100 .

Output

For each test case , print the answer mod 10000000071000000007 in a single line .

样例输入

0110011100011001001100011110001110001110001010010111111110101010010011010000110100011001001111101011

1000111101111110110100001101001101010001111001001011110001111110101000011101000001011100001001011010

0100101100011110011100110110011100111100010010011001111110101111111000000110001110000110001100001110

1110010101010001000110100011101010001010000110001111111110101010000000001111001110110101110000010011

1000010011111110001101100000101001110100011000111010011111110110111010011111010110101111011111011011

样例输出

515649254

题目来源

ACM-ICPC 2017 Asia Xi'an

题意：题意说的很乱，简单来说就是输入5个长为100的‘0’ ‘1’字符串，要求从每一行中取出一个‘1’ 但是这5个‘1’得在不同的列，求有多少种取法？注意输出的结果乘以常数tmp=531192758

思路：我们可以用DP解决这个问题，dp[i][j]=(dp[i][j]+dp[i-1][j-1])%mod 保证当前行取得一个‘1’ 是在前一行的‘1’的后面，然后把这5行字符串用全排列颠倒顺序即可，把所有排列顺序下的dp值dp[5][100]求和即可

这题正解是状压DP复杂度O(n)=2^5*500, 我上面的DP复杂度O(n)=5!*500, 这题现场赛的时候大多数人是暴力过的，唉~ 当时我傻了，DP想了一半觉得不太可行，赛后又想了想可行，比赛的时候压力有点大，有点紧张，特别是最后半个小时。

代码如下：

#include <iostream>

#include <algorithm>

#include <stdio.h>

#include <cstring>

using namespace std;

typedef long long LL;

const LL mod=1e9+;

char s[][];

LL dp[][],ans;

void Backtrack(int t)

{

    if(t==)

    {

        memset(dp,,sizeof(dp));

        for(int i=;i<=;i++)

        {

            dp[][i]=dp[][i-];

            if(s[][i]=='') dp[][i]++;

        }

        for(int i=;i<=;i++)

        {

            for(int j=;j<=;j++)

            {

                dp[i][j]=dp[i][j-];

                if(s[i][j]=='') dp[i][j]=(dp[i][j]+dp[i-][j-])%mod;

            }

        }

        ans=(ans+dp[][])%mod;

        return ;

    }

    for(int i=t;i<=;i++)

    {

        swap(s[i],s[t]);

        Backtrack(t+);

        swap(s[i],s[t]);

    }

}

int main()

{

    LL tmp=; /// 常数 tmp=A(95,5)*C(90,5)*C(85,5);

    while(scanf("%s",s[]+)!=EOF)

    {

        for(int i=;i<=;i++) scanf("%s",s[i]+);

        ans=;

        Backtrack();

        printf("%lld\n",ans*tmp%mod);

    }

    return ;

}

秒客网