Bzoj: [HNOI2005] Comb DP

: [HNOI2005] Comb time limit:10 Sec Memory limit:162 MB
submit:266 solved:125
[Submit] [Status] Description

Input

The first behavior of the integer L, wherein 4<=l<=100000, and 50% of the data to meet the l<=104, indicating the length of the lower side of the plank line segment. Second behavior L A positive integer a1,a2,..., AL, of which 1

Output

Contains only an integer d, which is the number of squares that need to be dug out of the board to make the comb area the largest.

Sample Input9
4 4 6 5 4 2 3 3 5Sample Output3
The general greedy strategy still needs a more rigorous proof. For example, this problem, position pos at the height of the possible should be h[i]+/-1 (pos-2<=i<=pos+2) A slight negligence, will be the scope of I wrong.

#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<vector>#include<map>using namespacestd;#defineMAXN 110000#defineINFL 0X3F3F3F3F3F3F3F3FLLtypedefLong LongQword;intH[MAXN];intpp[maxn][ +];; Qword dp[maxn][2][ +];inlinevoidDeal (Qword &X,qword y) {        if(x<y) x=y;}intMain () {Freopen ("Input.txt","R", stdin); inti,j,k,k2,x,y,z,n,m; Qword sum=0; scanf ("%d",&N);  for(i=1; i<=n;i++) {scanf ("%d", H +i); Sum+=H[i];  for(j=h[i]-1; j<=h[i]+1; j + +) {pp[i][++pp[i][0]]=J; if(I-1>=1) pp[i-1][++pp[i-1][0]]=J; if(i+1<=n) pp[i+1][++pp[i+1][0]]=J; if(I-2>=1) pp[i-2][++pp[i-2][0]]=J; if(i+2<=n) pp[i+2][++pp[i+2][0]]=J; }        }         for(i=1; i<=n;i++) {sort (Pp[i]+1, pp[i]+pp[i][0]+1); pp[i][0]=unique (&pp[i][1],&pp[i][pp[i][0]+1])-pp[i]-1;  for(j=pp[i][0]+1;j< A; j + +) pp[i][j]=0;  while(pp[i][0] && pp[i][pp[i][0]]>h[i]) pp[i][pp[i][0]--]=0; }         for(i=0; i<=n+1; i++)                 for(j=0;j<2; j + +)                         for(k=0;k< +; k++) Dp[i][j][k]=-INFL;  for(i=1; i<=pp[1][0];i++) dp[1][0][i]=dp[1][1][i]=pp[1][i];  for(i=2; i<=n;i++)        {                 for(j=1; j<=pp[i-1][0];j++)                {                         for(k=1; k<=pp[i][0];k++)                        {                                if(pp[i-1][j]<Pp[i][k]) {Deal (dp[i][0][k],dp[i-1][1][j]+Pp[i][k]); }Else if(pp[i-1][j]>Pp[i][k]) {Deal (dp[i][1][k],dp[i-1][0][j]+Pp[i][k]); }Else{deal (dp[i][0][k],dp[i-1][0][j]+Pp[i][k]); Deal (dp[i][1][k],dp[i-1][1][j]+Pp[i][k]); } }}} Qword ans=-INFL;  for(i=1; i<=pp[n][0];i++) {ans=max (ans,dp[n][0][i]); Ans=max (ans,dp[n][1][i]); } printf ("%lld", sum-ans);}

Bzoj: [HNOI2005] Comb DP