Problem Description
FSF is addicted to a stupid tower defense game. The goal of tower defense games is to try to stop enemies from crossing a map by building traps to slow them down and towers which shoot at them as they pass.
The map is a line, which has n unit length. We can build only one tower on each unit length. The enemy takes t seconds on each unit length. And there are 3 kinds of tower in this game: The red tower, the green tower and the blue tower.
The red tower damage on the enemy x points per second when he passes through the tower.
The green tower damage on the enemy y points per second after he passes through the tower.
The blue tower let the enemy go slower than before (that is, the enemy takes more z second to pass an unit length, also, after he passes through the tower.)
Of course, if you are already pass through m green towers, you should have got m*y damage per second. The same, if you are already pass through k blue towers, the enemy should have took t + k*z seconds every unit length.
FSF now wants to know the maximum damage the enemy can get.
Input
There are multiply test cases.
The first line contains an integer T (T<=100), indicates the number of cases.
Each test only contain 5 integers n, x, y, z, t (2<=n<=1500,0<=x, y, z<=60000,1<=t<=3)
Output
For each case, you should output "Case #C: " first, where C indicates the case number and counts from 1. Then output the answer. For each test only one line which have one integer, the answer to this question.
Sample Input
1 2 4 3 2 1
Sample Output
Case #1: 12 最优情况下,红塔必然都在最后,dp[i][j]代表前i个有j个蓝塔,枚举最后红塔的个数和前面蓝塔的个数进行dp即可#include<stdio.h> #include<string.h> #include<algorithm> #include<math.h> using namespace std; #define up(i,x,y) for(i=x;i<=y;i++) #define down(i,x,y) for(i=x;i>=y;i--) #define mem(a,b) memset(a,b,sizeof(a)) #define w(x) while(x) #define ll __int64 ll n,x,y,z,t,dp[1505][1505],ss,cas=1,i,j,k,r,ans; int main() { scanf("%I64d",&ss); w(ss--) { scanf("%I64d%I64d%I64d%I64d%I64d",&n,&x,&y,&z,&t); mem(dp,0); ans=n*t*x;//都是红塔 up(i,1,n) up(j,0,i) { if(!j) dp[i][j]=dp[i-1][j]+t*(i-j-1)*y; else dp[i][j]=max(dp[i-1][j]+(j*z+t)*max(0LL,(i-1-j))*y,dp[i-1][j-1]+((j-1)*z+t)*(i-j)*y); ans=max(ans,dp[i][j]+(n-i)*(j*z+t)*(x+(i-j)*y)); } printf("Case #%I64d: %I64d\n",cas++,ans); } return 0; }
HDU4939:Stupid Tower Defense(DP)