标签：als BMI space show eve pass others oss func
Stupid Tower Defense
Time Limit: 12000/6000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2504 Accepted Submission(s):
712
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
析：有一条长n段的路径，为了防御怪兽，需要在每一段路上建塔，塔的类型有3种：
1.红塔 对当前每个经过的怪兽造成x点伤害
2.绿塔 对经过绿塔之后的每个位置的怪兽都造成y值伤害
3.蓝塔 延迟怪兽在每一段路上花费z个时间，若前面经过k个蓝塔，则经过当前塔需要t+k*z时间
由于红塔为及时伤害，而绿塔为延后伤害，故可知红塔一定放在最后，才能造成更多的伤害，故假设全部建红塔, 用dp[i][j] 表示前i座塔中有j座蓝塔
第i座塔建绿塔: g = dp[i-1][j-1] + (i-j-1)*(t+z*j)*y
第i座塔建蓝塔: b = dp[i-1][j] + (i-j)*(t+z*(j-1))*y
dp[i][j] = max(g, b) , 再加上前面绿塔以及后面红塔对后面的影响即为当前最值
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <functional>
#define ll long long
#define ull unsigned ll
#define Inf 0x7fffffff
#define maxn 1000005
#define maxm 100005
#define P pair<int, int>
#define eps 1e-6
#define mod 10000
using namespace std;
const int N = 15005;
int T, n, m, len, ans;
ll t, x, y, z, dp[N][N];
int main(){
ans = 1;
scanf("%d", &T);
while(T--){
scanf("%d%lld%lld%lld%lld", &n, &x, &y, &z, &t);
ll res = x*t*n;
for(ll i = 1; i <= n; i ++){
for(ll j = 0; j <= i; j ++){
if(!j){
dp[i][j] = dp[i-1][j]+(i-j-1)*(t+j*z)*y; //建绿塔
}else{ // 前i座塔中有j座蓝塔
ll g = dp[i-1][j]+(i-j-1)*(t+j*z)*y; //第i座建绿塔
ll b = dp[i-1][j-1]+(i-j)*(t+(j-1)*z)*y;//第i座建蓝塔
dp[i][j] = max(g, b);
}
res = max(res, dp[i][j]+(y*(i-j)+x)*(t+z*j)*(n-i)); //加上后面红塔以及前面绿塔的影响
}
}
printf("Case #%d: %lld\n", ans++, res);
}
return 0;
}
View Code
hdu 4939
标签：als BMI space show eve pass others oss func
原文地址：https://www.cnblogs.com/microcodes/p/12787829.html
