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2017 ICPC区域赛（西安站）--- J题 LOL（DP）

2018-06-17 21:21:22来源：未知阅读 ()

题目链接

problem description

$5 friends play LOL together . Every one should BAN one character and PICK one character . The enemy should BAN$

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

Enemies pick heroes

Input

The input contains multiple test cases.(No more than

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

Output

For each test case , print the answer mod

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+7;
char s[6][105];
LL dp[6][105],ans;

void Backtrack(int t)
{
    if(t==5)
    {
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=100;i++)
        {
            dp[1][i]=dp[1][i-1];
            if(s[1][i]=='1') dp[1][i]++;
        }
        for(int i=2;i<=5;i++)
        {
            for(int j=1;j<=100;j++)
            {
                dp[i][j]=dp[i][j-1];
                if(s[i][j]=='1') dp[i][j]=(dp[i][j]+dp[i-1][j-1])%mod;
            }
        }
        ans=(ans+dp[5][100])%mod;
        return ;
    }
    for(int i=t;i<=5;i++)
    {
        swap(s[i],s[t]);
        Backtrack(t+1);
        swap(s[i],s[t]);
    }
}

int main()
{
    LL tmp=531192758; /// 常数 tmp=A(95,5)*C(90,5)*C(85,5);
    while(scanf("%s",s[1]+1)!=EOF)
    {
        for(int i=2;i<=5;i++) scanf("%s",s[i]+1);
        ans=0;
        Backtrack(1);
        printf("%lld\n",ans*tmp%mod);
    }
    return 0;
}