D. Nastya and Scoreboard
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Denis, after buying flowers and sweets (you will learn about this story in the next task), went to a date with Nastya to ask her to become a couple. Now, they are sitting in the cafe and finally... Denis asks her to be together, but ... Nastya doesn't give any answer.
The poor boy was very upset because of that. He was so sad that he punched some kind of scoreboard with numbers. The numbers are displayed in the same way as on an electronic clock: each digit position consists of 77 segments, which can be turned on or off to display different numbers. The picture shows how all 1010 decimal digits are displayed:
After the punch, some segments stopped working, that is, some segments might stop glowing if they glowed earlier. But Denis remembered how many sticks were glowing and how many are glowing now. Denis broke exactly kk segments and he knows which sticks are working now. Denis came up with the question: what is the maximum possible number that can appear on the board if you turn on exactly kk sticks (which are off now)?
It is allowed that the number includes leading zeros.
Input
The first line contains integer nn (1≤n≤2000)(1≤n≤2000) — the number of digits on scoreboard and kk (0≤k≤2000)(0≤k≤2000) — the number of segments that stopped working.
The next nn lines contain one binary string of length 77, the ii-th of which encodes the ii-th digit of the scoreboard.
Each digit on the scoreboard consists of 77 segments. We number them, as in the picture below, and let the ii-th place of the binary string be 00 if the ii-th stick is not glowing and 11 if it is glowing. Then a binary string of length 77 will specify which segments are glowing now.
Thus, the sequences "1110111", "0010010", "1011101", "1011011", "0111010", "1101011", "1101111", "1010010", "1111111", "1111011" encode in sequence all digits from 00 to 99 inclusive.
Output
Output a single number consisting of nn digits — the maximum number that can be obtained if you turn on exactly kk sticks or −1−1, if it is impossible to turn on exactly kk sticks so that a correct number appears on the scoreboard digits.
Examples
input
Copy
1 7 0000000
output
Copy
8
input
Copy
2 5 0010010 0010010
output
Copy
97
input
Copy
3 5 0100001 1001001 1010011
output
Copy
-1
Note
In the first test, we are obliged to include all 77 sticks and get one 88 digit on the scoreboard.
In the second test, we have sticks turned on so that units are formed. For 55 of additionally included sticks, you can get the numbers 0707, 1818, 3434, 4343, 7070, 7979, 8181 and 9797, of which we choose the maximum — 9797.
In the third test, it is impossible to turn on exactly 55 sticks so that a sequence of numbers appears on the scoreboard.
題意：
每個數字由7根顯示管組成，按順序給出每個顯示屏現在的狀態，0表示顯示管關閉，1表示打開；現在從原有的基礎上精確地打開 k 個顯示管，問是否可以顯示出一串數字，如果可以，求出能顯示的最大的數字
思路：
（我沒有想到dp，爆搜T了……
dp：
dp[ i ][ j ]表示從第 i 個數字開始到最後，點亮 j 個顯示管，能否顯示出數字。（從後向前）
現預處理出原始的每個顯示屏變到某個數字需要點亮的顯示管數，或者不可以變成某個數字
貪心時從前向後，從9到0（保證數字儘量大）
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int inf = 0x3f3f3f3f;
const double eps = 1e-8;
const double PI = acos(-1.0);
const int N = 2e3 + 10;
string ss[10] = {"1110111", "0010010", "1011101", "1011011", "0111010", "1101011", "1101111", "1010010", "1111111", "1111011"};
string s[N];
int n, k, cn[N][10];
bool dp[N][N];
void init()
{
memset(cn, 0, sizeof(cn));
for(int i = 1; i <= n; ++i)
{
for(int j = 0; j < 10; ++j)
{
for(int k = 0; k < 7; ++k)
{
if(s[i][k] == '1' && ss[j][k] == '0')
{
cn[i][j] = -1;
break;
}
if(s[i][k] != ss[j][k])
{
cn[i][j]++;
}
}
// cout<<cn[i][j]<<' ';
}
// cout<<'\n';
}
}
bool solve()
{
memset(dp, 0, sizeof(dp));
dp[n + 1][0] = 1;
for(int i = n; i; --i)
{
for(int j = 0; j < 10; ++j)
{
if(cn[i][j] != -1)
{
for(int u = cn[i][j]; u <= k; ++u)
{
dp[i][u] |= dp[i + 1][u - cn[i][j]];
}
}
}
}
if(!dp[1][k])
return 0;
else
return 1;
}
int main()
{
while(~scanf("%d%d", &n, &k))
{
for(int i = 1; i <= n; ++i)
{
cin >> s[i];
}
init();
if(!solve())
cout<<-1<<'\n';
else
{
int tmp = k;
vector<int>vec;
for(int i = 1; i <= n; ++i)
{
for(int j = 9; j >= 0; --j)
{
if(cn[i][j] != -1 && tmp >= cn[i][j] && dp[i + 1][tmp - cn[i][j]])
{
tmp -= cn[i][j];
vec.push_back(j);
break;
}
}
}
for(int i = 0; i < vec.size(); ++i)
{
cout<<vec[i];
}
cout<<'\n';
vec.clear();
}
}
return 0;
}
dfs剪枝：
等會來補