hdu4939 Stupid Tower Defense （DP）

2014多校7 第二水的题

4939

Stupid Tower Defense

Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 366    Accepted Submission(s): 88

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

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.

 

Author

UESTC

 

Source

2014 Multi-University Training Contest 7

 

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题意：塔防，怪过每个块用的初始时间为t。每块可以建一个塔，有三种塔，一种是红塔，怪经过当前格子时每秒输出x；一种绿塔，怪过了当前格子后每秒被毒y血；一种是蓝塔，怪经过当前格子后经过每格的时间增加z秒。绿塔和蓝塔效果都可叠加，求最高输出。

题解：DP。

思考题面，可以想到红塔放最后是最优解，可以反证得：如果有个红塔后面有个不红的塔，交换它们肯定可以得更优解。

这样我们就枚举不红的塔的数量、绿塔的数量，f[i][j]表示有前面有i个不红的塔，其中绿塔有j个时，蓝绿塔在全路段的输出和。

因为已知i,j时，后面的红塔的输出也是固定的（只随着i,j变化，ij固定时红塔输出不变），所以我们dp求得各种ij的最大的f[i][j]，ans每次更新。

由于数好像有点大，用超碉的会滚的队列比较爽。

 //#pragma comment(linker, "/STACK:102400000,102400000")
 #include<cstdio>
 #include<cmath>
 #include<iostream>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 #include<map>
 #include<set>
 #include<stack>
 #include<queue>
 using namespace std;
 #define ll long long
 #define usll unsigned ll
 #define mz(array) memset(array, 0, sizeof(array))
 #define minf(array) memset(array, 0x3f, sizeof(array))
 #define REP(i,n) for(i=0;i<(n);i++)
 #define FOR(i,x,n) for(i=(x);i<=(n);i++)
 #define RD(x) scanf("%d",&x)
 #define RD2(x,y) scanf("%d%d",&x,&y)
 #define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)
 #define WN(x) prllf("%d\n",x);
 #define RE  freopen("D.in","r",stdin)
 #define WE  freopen("1biao.out","w",stdout)
 ll max(ll x,ll y) {
     return x>y?x:y;
 }
 ll f[][][];///green blue

 int main() {
     ll T,cas=;
     ll n,t;
     ll x,y,z;
     ll now,pre;
     ll i,j,k;
     scanf("%I64d",&T);
     while(T--) {
         scanf("%I64d%I64d%I64d%I64d%I64d",&n,&x,&y,&z,&t);
         memset(f,,sizeof(f));
         now=;
         pre=;
         ll ans=x*t*n;
         for(i=; i<=n; i++) { ///第1到第i个不放红的
             for(j=; j<=i; j++) { ///第1个到第i个有j个green
                 k=i-j;///第1个到第i个有k个blue
                 if(k>) {///第i个放蓝的
                     f[now][j][k]=f[pre][j][k-] + j*y*z*(n-i);///放这个蓝的接下来n-i格每格增加了z时间
                 }
                 if(j>) {///放绿的
                     f[now][j][k]=max(f[now][j][k],f[pre][j-][k] + y*(t+k*z)*(n-i) );///放这个绿的每秒增加y伤害
                 }
                 ans=max(ans,f[now][j][k] + x*(n-i)*(t+k*z));
             }
             pre=now;
             now^=;
         }
         printf("Case #%I64d: %I64d\n",cas++,ans);
     }
     return ;
 }