# | Time | Username | Problem | Language | Result | Execution time | Memory |
---|---|---|---|---|---|---|---|
55781 | okaybody10 | Miners (IOI07_miners) | C++98 | 120 ms | 1564 KiB |
This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include <bits/stdc++.h>
using namespace std;
const int mn=-1e9;
int dp[2][4][4][4][4],cost[4][4][4]; // cost[i][j][k] = 음식이 i,j,k 순으로 들어왔을 경우 광부들로부터 얻을 수 있는 석탄의 량
int change[100007],n;
char s[100008];
int main()
{
scanf("%d %s",&n,s);
for(int i=0;i<4;i++)
{
for(int j=0;j<4;j++)
{
for(int k=0;k<3;k++)
{
if(i==3) // 비어있는 경우 : 3, M : 0, B : 1, F : 2
{
if(j==3) cost[i][j][k]=1; // 두 개다 비어있으므로 k의 종류에 상관없이 무조건 1
else cost[i][j][k]=1+(j!=k ? 1 : 0); // 한 개는 비어있지 않으므로 j랑 k가 동일한 종류라면 1, 그 외엔 2
}
else if(j==3)
{
cost[i][j][k]=mn;
}
else
{
if(i!=j && j!=k && i!=k) cost[i][j][k]=3;
else if(i==j && j==k) cost[i][j][k]=1;
else cost[i][j][k]=2;
}
}
}
}
for(int i=0;i<n;i++)
{
if(s[i]=='M') change[i]=0;
else if(s[i]=='B') change[i]=1;
else change[i]=2;
}
for(int i=0;i<4;i++) for(int j=0;j<4;j++) for(int k=0;k<4;k++) for(int l=0;l<4;l++) dp[0][i][j][k][l]=mn;
dp[0][3][3][3][3]=0;
for(int i=0;i<n;i++)
{
for(int j=0;j<4;j++) for(int k=0;k<4;k++) for(int l=0;l<4;l++) for(int v=0;v<4;v++) dp[(i+1)%2][j][k][l][v]=mn;
for(int j=0;j<4;j++) for(int k=0;k<4;k++) for(int l=0;l<4;l++) for(int v=0;v<4;v++)
{
if(dp[i%2][j][k][l][v]>=0)
{
dp[(i+1)%2][j][k][v][change[i]]=max(dp[(i+1)%2][j][k][v][change[i]],dp[i%2][j][k][l][v]+cost[l][v][change[i]]);
dp[(i+1)%2][k][change[i]][l][v]=max(dp[(i+1)%2][k][change[i]][l][v],dp[i%2][j][k][l][v]+cost[j][k][change[i]]);
}
}
}
int ans=mn;
for(int i=0;i<4;i++) for(int j=0;j<4;j++) for(int k=0;k<4;k++) for(int l=0;l<4;l++) ans=max(ans,dp[n%2][i][j][k][l]);
return !printf("%d",ans);
}
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