2014多校7 第二水的题
Stupid Tower DefenseTime 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
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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 ;
}