ACM-ICPC 2017 Asia Xi'an J LOL 【暴力 && 排列组合】

任意门：https://nanti.jisuanke.com/t/20750

J - LOL

5 friends play LOL together . Every one should BAN one character and PICK one character . The enemy should BAN $5$ characters and PICK $5$ characters . All these $20$ 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 $1$ : picks hero $1$, bans hero $2$

Player $2$ : picks hero $3$, bans hero $4$

Player $3$ : picks hero 5, bans hero $6$

Player $4$ : picks hero $7$, bans hero $8$

Player $5$ : picks hero $9$, bans hero $10$

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

Input

The input contains multiple test cases.(No more than $20$)

In each test case . there’s $5$ strings $S[1]\sim S[5]$ ,respectively whose lengths are $100$ , For the $i$-th person if he has bought the $j$-th hero, the $j$-th character of $S[i]$ is '$1$', or '$0$' if not. The total number of heroes is exactly $100$ .

Output

For each test case , print the answer mod $1000000007$ in a single line .

样例输入复制

0110011100011001001100011110001110001110001010010111111110101010010011010000110100011001001111101011
1000111101111110110100001101001101010001111001001011110001111110101000011101000001011100001001011010
0100101100011110011100110110011100111100010010011001111110101111111000000110001110000110001100001110
1110010101010001000110100011101010001010000110001111111110101010000000001111001110110101110000010011
1000010011111110001101100000101001110100011000111010011111110110111010011111010110101111011111011011

样例输出复制

515649254

题意概括：

五个二进制数代表我方 5 人所拥有的英雄，0为没有，1为有。

敌方5人什么英雄都有，问有多少种 Ban Pick方案。

每人都可以选一个自己拥有的英雄和 ban 掉任何一个英雄，最后选的和禁止掉的20个英雄都不相同。

解题思路：

暴力枚举我方选英雄的方案

敌方选英雄的方案为 A（95， 5）

我方禁英雄的方案为 C（90， 5）

敌方禁英雄的方案为 C（85，5）

关于暴力的优化和剪枝：

如果是传统暴力枚举每个人选不同英雄的方案，需要五层 for（0， 100） 循环，TL！

优化只枚举前四个人的方案， 第五个能选的英雄只能是前面没有选的，缩小了一层循环。

关于剪枝 前面选过的不要选咯

 

AC code：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#define INF 0x3f3f3f3f
#define LL long long
using namespace std;
const int mod = 1e9+7;
const int MAXN = 105;

char a[MAXN], b[MAXN], c[MAXN], d[MAXN], e[MAXN];
LL A(LL N, LL R)
{
    LL sum = 1;
    for(int i = 0; i < R; i++)
        sum *= N-i;
    return sum;
}

LL C(LL N, LL R)
{
    LL sum = 1;
    for(int i = 1; i <= R; i++)
        sum = sum* (N+1-i)/i;
    return sum;
}

int main()
{
    while(~scanf("%s%s%s%s%s", a, b, c, d, e)){
        int len = 0;
        for(int i = 0; i < 100; i++){
            if(e[i] == '1') len++;
        }
        LL ans = 0;
        for(int i = 0; i < 100; i++){
            if(a[i] == '0') continue;
            for(int j = 0; j < 100; j++){
                if(b[j] == '0' || j == i) continue;
                for(int k = 0; k < 100; k++){
                    if(c[k] == '0' || k == i || k == j) continue;
                    for(int p = 0; p < 100; p++){
                        if(d[p] == '0' || p == i || p == j || p == k) continue;
                        int res = len;
                        if(e[i] == '1') res--;
                        if(e[j] == '1') res--;
                        if(e[k] == '1') res--;
                        if(e[p] == '1') res--;
                        if(res>=0)      ans+=res;
                    }
                }
            }
        }
        ans = ans*A(95, 5)%mod*C(90, 5)%mod*C(85, 5)%mod;
        printf("%lld\n", ans);
    }
    return 0;
}