Tc srm 551 div2

Reprint please indicate the source, thank you http://blog.csdn.net/ACM_cxlove? Viewmode = Contents
By --- cxlove

I finally passed two questions in TC. Unfortunately, the third question was wrong.

250pt: only several colors are required.

class ColorfulBricks{public:int countLayouts(string bricks){bool flag[26];int cnt=0;memset(flag,false,sizeof(flag));for(int i=0;i<bricks.size();i++)if(!flag[bricks[i]-'A']){flag[bricks[i]-'A']=true;cnt++;}if(cnt==1)return 1;if(cnt==2)return 2;return 0;}};

500pt

class ColorfulChocolates{public:int maximumSpread(string chocolates, int maxSwaps){int mmax=1;int len=chocolates.size();for(int i=0;i<26;i++){for(int j=0;j<len-1;j++){int a=j,b=j+1;int tmp=0;int step=maxSwaps;bool flag[60];memset(flag,true,sizeof(flag));for(int x=a,y=b;x>=0||y<len;x--,y++){if(x>=0&&x<len&&chocolates[x]==i+'A'){if(step>=a-x){tmp++;step-=a-x;a--;}}if(y>=0&&y<len&&chocolates[y]==i+'A'){if(step>=y-b){tmp++;step-=(y-b);b++;}    }}mmax=max(mmax,tmp);}}return mmax;}};

1000pt: 26 colors are used in the game, and the polyA count using multiple sets and matrices is useless. There are only three colors, and the DP can be broken. I wrote two versions, and the first one was tested on the local machine. The traffic was not mentioned, but the memory should be too large.

Note: DP [POS] [a] [B] [C] [kind] indicates the position of the previous pos. a A, B, and C are used. The last one is the kind number. Enumeration start point.

Type 2: DP [FST] [pre] [a] [B] [C] indicates that the first one is FST, and the last one is pre. a A is used, number of B and C types. Memory-based search can be broken.

int dp[3][3][51][51][51];int cnt[3],n;class ColorfulCupcakesDivTwo{public:int slove(int pos,int pre,int fst,int a,int b,int c){if(a<0||b<0||c<0)return 0;if(pos==n)return pre!=fst;if(dp[fst][pre][a][b][c]!=-1)return dp[fst][pre][a][b][c];int ret=0;if(pos==0){ret=(ret+slove(pos+1,0,0,a-1,b,c))%MOD;ret=(ret+slove(pos+1,1,1,a,b-1,c))%MOD;ret=(ret+slove(pos+1,2,2,a,b,c-1))%MOD;}else if(pre==0){ret=(ret+slove(pos+1,fst,1,a,b-1,c))%MOD;ret=(ret+slove(pos+1,fst,2,a,b,c-1))%MOD;}else if(pre==1){ret=(ret+slove(pos+1,fst,0,a-1,b,c))%MOD;ret=(ret+slove(pos+1,fst,2,a,b,c-1))%MOD;}else if(pre==2){ret=(ret+slove(pos+1,fst,0,a-1,b,c))%MOD;ret=(ret+slove(pos+1,fst,1,a,b-1,c))%MOD;}return dp[fst][pre][a][b][c]=ret;}int countArrangements(string cupcakes){cnt[0]=cnt[1]=cnt[2]=0;n=cupcakes.size();for(int i=0;i<n;i++)cnt[cupcakes[i]-'A']++;memset(dp,-1,sizeof(dp));return slove(0,0,0,cnt[0],cnt[1],cnt[2]);}};/*int dp[51][51][51][51][3];int countArrangements(string cupcakes){cnt[0]=cnt[1]=cnt[2]=0;int n=cupcakes.size();for(int i=0;i<n;i++)cnt[cupcakes[i]-'A']++;int ans=0;for(int s=0;s<3;s++){memset(dp,0,sizeof(dp));if(s==0)dp[1][1][0][0][0]=1;else if(s==1)dp[1][0][1][0][1]=1;elsedp[1][0][0][1][2]=1;for(int x=2;x<=n;x++){for(int a=0;a<=cnt[0];a++)for(int b=0;b<=cnt[1];b++)for(int c=0;c<=cnt[2];c++)if(a+b+c==x){if(a){dp[x][a][b][c][0]=(dp[x][a][b][c][0]+dp[x-1][a-1][b][c][1])%MOD;dp[x][a][b][c][0]=(dp[x][a][b][c][0]+dp[x-1][a-1][b][c][2])%MOD;}if(b){dp[x][a][b][c][1]=(dp[x][a][b][c][1]+dp[x-1][a][b-1][c][0])%MOD;dp[x][a][b][c][1]=(dp[x][a][b][c][1]+dp[x-1][a][b-1][c][2])%MOD;}if(c){dp[x][a][b][c][2]=(dp[x][a][b][c][2]+dp[x-1][a][b][c-1][1])%MOD;dp[x][a][b][c][2]=(dp[x][a][b][c][2]+dp[x-1][a][b][c-1][0])%MOD;}}}ans+=(((-dp[n][cnt[0]][cnt[1]][cnt[2]][s]+dp[n][cnt[0]][cnt[1]][cnt[2]][0]+dp[n][cnt[0]][cnt[1]][cnt[2]][2]))+dp[n][cnt[0]][cnt[1]][cnt[2]][1])%MOD;    cout<<ans<<endl;}return ans;}*/