ACM-ICPC 2017 Asia Xi'an LOL dfs剪枝|dp

题目链接

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

/*
0110011100011001001100011110001110001110001010010111111110101010010011010000110100011001001111101011
1000111101111110110100001101001101010001111001001011110001111110101000011101000001011100001001011010
0100101100011110011100110110011100111100010010011001111110101111111000000110001110000110001100001110
1110010101010001000110100011101010001010000110001111111110101010000000001111001110110101110000010011
1000010011111110001101100000101001110100011000111010011111110110111010011111010110101111011111011011
*/
#include<algorithm>
#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<string.h>
using namespace std;
char a[10][222];
long long sum,ans;
const long long mod = 1e9+7;
int book[222];
int dis[222];
void dfs(int x)
{
if(x==5)
{
ans=(sum+ans);
for(int i=1;i<=4;i++)
{
if(a[5][dis[i]]=='1')
ans--;
}

ans = ans%mod;
return ;
}
for(int i=0;i<100;i++)
{
if(book[i]==0&&a[x][i]=='1')
{
book[i] = 1;
dis[x] = i;
dfs(x+1);
book[i] = 0;
}
}
}
int main()
{
while(cin>>a[1])
{
for(int i=2;i<=5;i++)
{
cin>>a[i];
}
ans = 0;
sum = 0;
memset(book,0,sizeof book);
for(int i=0;i<100;i++)
if(a[5][i]=='1')sum++;
dfs(1);
ans = (ans*531192758)%mod;
cout<<ans<<endl;
}
return 0;
}

dp版o( 5！* 500)

参考博客

code:

#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<string.h>
using namespace std;
char a[10][222];
long long sum,ans;
const long long mod = 1e9+7;
int book[222];
int dis[222];
long long dp[10][222];
void dfs(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(a[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(a[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(a[i],a[t]);
dfs(t+1);
swap(a[i],a[t]);
}
}
int main()
{
long long tmp = 531192758;
while(~scanf("%s",a[1]+1))
{
for(int i=2;i<=5;i++)
{
scanf("%s",a[i]+1);
}
ans = 0;
sum = 0;
memset(book,0,sizeof book);
dfs(1);
ans = (ans*tmp)%mod;
cout<<ans<<endl;
}
return 0;
}