标签：des style blog color io os ar java for
Stupid Tower Defense
Time Limit: 12000/6000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1589 Accepted Submission(s): 452
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
Sample Output
Case #1: 12
Hint
For 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.
题目给出一条长度为n的直线 ， 你可以在长度为1的单元上面建造塔。
有三种类型的塔
颜色 效果
red 经过这个塔的时候，收到 x (d / s) 的伤害
green 经过这个塔之后 ， 受到 y (d / s) 的持续伤害
bule 经过这个塔之后 ， 经过1个塔的时间增加 z 秒
题目要求 ， 求出在直线上放置n个塔 ， 使得经过直线后受到的伤害最大。
然后可想而知， 红色无论放多少个。 最后放肯定是没错的 ， 因为能够得到蓝色的叠加时间 ， 伤害效果尽量大。
剩下就是蓝色跟绿色 ， 应该怎么放。
这时候就要用到DP了 。
dp[i][j] 。。。表示前 i + j (<=n ) 个长度 i 个放green ， j 个放blue 所得到的最大伤害 。
转移就是这样取就好了
dp[i][j] = max( dp[i-1][j] + ( j * z + t ) * y * ( i - 1 ) , dp[i][j-1] + ( ( j - 1 ) * z + t ) * y * i );
注意一下边界 。
然后最后要补上 red 塔的伤害就得出答案了~
#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;
typedef long long ll;
const int N = 1550 ;
ll dp[N][N]; // i = green , j = blue ...
int main()
{
ll _ , n , x , y , z , t , cas = 1;
#ifdef LOCAL
freopen("in","r",stdin);
#endif
cin >> _ ;
while( _ -- ){
cin >> n >> x >> y >> z >> t ;
cout << "Case #"<< cas++ <<": ";
memset(dp,0,sizeof dp);
for( int i = 1 ; i <= n ; ++i ){
for( int j = 0 ; j <= i ; ++j ){
int k = i - j ;
ll temp = 0 ;
if( j > 0 ) temp = max( temp , dp[j-1][k] + ( k * z + t ) * y * ( j - 1 ) );
if( k > 0 ) temp = max( temp , dp[j][k-1] + ( ( k - 1 ) * z + t ) * y * j ) ;
dp[j][k] = temp ;
}
}
ll ans = 0 ;
for( int i = 0 ; i <= n ; ++i ){
for( int j = 0 ; j + i <= n ; ++j ){
ans = max ( ans , dp[i][j] + ( n - i - j ) * ( x * ( t + z * j ) + i * y * ( t + z * j ) ) );
}
}
cout << ans << endl ;
}
return 0;
}
hdu 4939 Stupid Tower Defense
标签：des style blog color io os ar java for
原文地址：http://www.cnblogs.com/hlmark/p/4014784.html
