ACM Learning Process-Guangdong University of Technology 2016 final-online game e block water (maximum problem | | Dynamic planning)

Title Link:http://gdutcode.sinaapp.com/problem.php?cid=1031&pid=4

This topic will naturally be considered to discuss the longest or shortest board.

The pen may be simulated, you will know, assuming the longest board isR,0and then+1The location is high for0of the board, then for[0, R-1]The longest board in theRR,RRto theRThis short should be a depth ofA[RR]of water cover. The same[0, Rr-1]The longest board in theRRR,RRRto theRRThis paragraph is supposed to beA[rrr]cover, and so on can be doneRthe first paragraph, the same as the latter paragraph.

With regard to the maximum value of this piece, you can use logn maintenance such as RMQ , the total complexity is nlogn. But considering that the intervals are [0, R] and [R, N+1] this. So it is easy to think of DP,p[0][i] represents the maximum value from 0 to I, p[1][i] represents the angle of the maximum value from I to the n+1 interval. And then two times the equation shifts. The total complexity is O (n).

Code:

#include <iostream>#include<cstdio>#include<cstdlib>#include<cmath>#include<cstring>#include<algorithm>#include<Set>#include<map>#include<queue>#include<vector>#include<string>#defineLL Long Longusing namespacestd;Const intMAXN = 1e6+5;intN, A[MAXN];//rmq-st Algorithm//Efficiency Nlogn//Sany, note the difference between the interval [0, n-1] and [1, N]intma[maxn][ -];voidRMQ () {memset (MA,0,sizeof(MA));  for(inti =0; I <= n+1; ++i) ma[i][0] =i;  for(intj =1; (1&LT;&LT;J) <= n+2; ++j) for(inti =0; i+ (1&LT;&LT;J)-1<= n+1; ++i) {if(a[ma[i][j-1]] > a[ma[i+ (1<< (J-1))][j-1] ]) ma[i][j]= ma[i][j-1]; ElseMa[i][j]= ma[i+ (1<< (J-1))][j-1]; }}intQueryintLtintRT) {    intK =0;  while((1<< (k +1)) <= rt-lt+1) K++; if(A[ma[lt][k]] > a[ma[rt-(1&LT;&LT;K) +1][k]])returnMa[lt][k]; Else        returnMa[rt-(1&LT;&LT;K) +1][k];}voidinput () {scanf ("%d", &N); a[0] = a[n+1] =0;  for(inti =1; I <= N; ++i) scanf ("%d", &A[i]); RMQ ();} LL Cal (intLtintRtinth) {LL ans=0;  for(inti = lt; I <= RT; ++i) ans+ = Max (0, H-A[i]); returnans;}voidWork () {LL ans=0; intMid = Query (0, n+1), K, now; now=mid;  while(Now >0) {k= Query (0, now-1); Ans+ = Cal (k, now-1, A[k]); now=K; } Now=mid;  while(Now < n+1) {k= Query (now+1, n+1); Ans+ = Cal (now+1, K, a[k]); now=K; } cout<< ans <<Endl;}intMain () {//freopen ("test.in", "R", stdin);    intT; scanf ("%d", &T);  for(intTimes =1; Times <= T; ++Times )        {input ();    Work (); }    return 0;}

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ACM Learning Process-Guangdong University of Technology 2016 final-online game e block water (maximum problem | | Dynamic planning)