标签：des style blog http color os io for
Stupid Tower Defense
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
题意： 给出3种塔防装置，有红塔（只经过的时候掉分），绿塔（经过之后每秒都掉分），兰塔 (能延缓怪物的行进速度)。 问如何安置塔防恩能使得
怪物掉分最多。
sl : 刚开始题意理解错了逗比了一个小时，后来知道， t + k*z 是固定的。 这就是傻逼的dp了。
dp[i][j]=max(dp[i-1][j-1]+(t+(j-1)*z)*y*(i-j),dp[i-1][j]+(t+j*z)*y*(i-j-1));
i表示当前是第i个位置。 j表示前面有多少个兰塔。因为红塔放在最后面肯定就是最优的（显而易见）。
1 // by caonima
2 // hehe
3 #include <cstdio>
4 #include <cstring>
5 #include <algorithm>
6 #include <vector>
7 #include <map>
8 using namespace std;
9 typedef
long long LL;
10 const int MAX=
2000+
10;
11 const LL inf = 1LL<<
15;
12 LL n,x,y,z,t;
13 LL dp[MAX][MAX];
14 LL gao() {
15 for(LL i=
1;i<=n;i++) {
16 for(LL j=
0;j<=i;j++) {
17 if(j>
0) dp[i][j]=dp[i-
1][j-
1]+(t+(j-
1)*z)*y*(i-j);
18 dp[i][j]=max(dp[i][j],dp[i-
1][j]+(t+j*z)*y*(i-j-
1));
19 }
20 }
21 LL ans=-inf;
22 for(LL i=
1;i<=n;i++) {
23 for(LL j=
0;j<=i;j++) {
// blue tower
24 ans=max(ans,dp[i][j]+(n-i)*x*(t+j*z)+(n-i)*(i-j)*(t+j*z)*y);
25 }
26 }
27 ans=max(ans,n*x*t);
28 return ans;
29 }
30 int main() {
31 int cas,cnt=
0;
32 scanf(
"%d",&cas);
33 while(cas--) {
34 35 scanf(
"%I64d %I64d %I64d %I64d %I64d",&n,&x,&y,&z,&t);
36 memset(dp,
0,
sizeof(dp));
37 LL ans=gao();
38 printf(
"Case #%d: %I64d\n",++cnt,ans);
39 }
40 }
41 42 /*
43 3
44 3 20 100 5 1
45
46 */
HDU 4939,布布扣,bubuko.com
HDU 4939
标签：des style blog http color os io for
原文地址：http://www.cnblogs.com/acvc/p/3908287.html
