Stupid Tower Defense
Time Limit: 12000/6000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1557 Accepted Submission(s): 445
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
HintFor the first sample, the first tower is blue tower, and the second is red tower. So, the total damage is 4*(1+2)=12 damage points.
Author UESTC
Source 2014 Multi-University Training Contest 7 分析：红塔只对过塔的人功击，绿塔对之后的所有塔的走过人功击，篮塔只会廷长过塔时间，假设有n个塔，有e个绿塔，j个篮塔，k个红塔，n=e+j+k;红塔不管放哪里都只对当前过塔人功击，要想总功击最大那么所有的红塔必须放在最后，那么现在只要在枚举出e和j的个数，k＝n-e-j; k个红塔都排在最后，设dp[i][j]表示前i个塔中有j个篮塔的最大功击值。
#include
#include
#define ll __int64 ll dp[1505][1505]; int main() { ll T,n,x,y,z,t,ans,c=0,aa; for(int i=0;i<=1500;i++) dp[0][i]=0; scanf("%I64d",&T); while(T--) { scanf("%I64d%I64d%I64d%I64d%I64d",&n,&x,&y,&z,&t); ans=n*t*x; for(ll i=1;i<=n;i++) for(ll j=0;j<=i;j++) { dp[i][j]=dp[i-1][j]+(i-1-j)*(j*z+t)*y;//第i个塔是绿塔 if(j>0) { aa=dp[i-1][j-1]+(i-j)*((j-1)*z+t)*y;//第i个塔是篮塔 if(dp[i][j]
ans) ans=aa; } printf("Case #%I64d: %I64d\n",++c,ans); } }