Bailian 4118: Open a restaurant

Bailian 4118: Open a restaurant

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Total time limit:1000ms Memory Limit:65536kBDescriptionXiao Ming, a student of PKU School of Information, intends to open a restaurant after graduation. There are now N locations to choose from. Xiao Ming intends to choose the right location to open some restaurants. The n locations are arranged in a straight line. We use an integer sequence m1, M2, ... mn to represent their relative position. Because of the lot relationship, the profit of the restaurant will be different. We use PI to show the profit of opening a restaurant in Mi. In order to avoid the internal competition of their own restaurants, the distance between restaurants must be greater than K. Please help Xiao Ming to choose a plan with the biggest total profit.  inputThe standard input contains several sets of test data. The input first line is the integer t (1 <= T <= 1000), indicating that there is a T group of test data. followed by a continuous test of the T-group. Each set of test data has 3 rows,Line 1th: Total number of places N (n < 100), distance limit K (k > 0 && K < +).Line 2nd: N Locations M1, M2, ... mn (1000000 > Mi > 0 and Integer, ascending)Line 3rd: Restaurant profit in n locations p1, p2, ... pn (> Pi > 0 and integer)Outputmaximum possible profit for each set of test dataSample Input2 3 1 2 (2 , 3, 1 2) 2Sample Output All Save the position with array A, array B holds the profit for each position, and the array C represents the total profit.  

#include <cstdio>#include<cstring>#include<iostream>#include<queue>#include<stack>#include<cmath>#include<map>#include<algorithm>using namespaceStd;map<string,string>Mp;vector<vector<int> >G;Const intinf=0x3f3f3f3f;Const intmaxn=1100;intA[MAXN], B[MAXN], C[MAXN];intN, M;intMain () {intT; scanf ("%d", &T);  while(t--) {scanf ("%d%d", &n, &m);  for(intI=1; i<=n; i++) scanf ("%d", &A[i]);  for(intI=1; i<=n; i++) {scanf ("%d", &B[i]); C[i]=B[i]; }          for(intI=1; i<=n; i++)        {             for(intj=1; j<=i-1; J + +)                if(a[i]-a[j]>m) C[i]=max (C[i], c[j]+B[i]); }         intans=0;  for(intI=1; i<=n; i++) ans=Max (ans, c[i]); printf ("%d\n", ans); }    return 0;}

 

Bailian 4118: Open a restaurant