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. Input There are multiply test cases. 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 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 Recommend We have carefully selected several similar problems for you: 4944 4943 4942 4941 4940 |
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。
1 //#pragma comment(linker, "/STACK:102400000,102400000")
2 #include<cstdio>
3 #include<cmath>
4 #include<iostream>
5 #include<cstring>
6 #include<algorithm>
7 #include<cmath>
8 #include<map>
9 #include<set>
10 #include<stack>
11 #include<queue>
12 using namespace std;
13 #define ll long long
14 #define usll unsigned ll
15 #define mz(array) memset(array, 0, sizeof(array))
16 #define minf(array) memset(array, 0x3f, sizeof(array))
17 #define REP(i,n) for(i=0;i<(n);i++)
18 #define FOR(i,x,n) for(i=(x);i<=(n);i++)
19 #define RD(x) scanf("%d",&x)
20 #define RD2(x,y) scanf("%d%d",&x,&y)
21 #define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)
22 #define WN(x) prllf("%d\n",x);
23 #define RE freopen("D.in","r",stdin)
24 #define WE freopen("1biao.out","w",stdout)
25 ll max(ll x,ll y) {
26 return x>y?x:y;
27 }
28 ll f[2][1555][1555];///green blue
29
30 int main() {
31 ll T,cas=1;
32 ll n,t;
33 ll x,y,z;
34 ll now,pre;
35 ll i,j,k;
36 scanf("%I64d",&T);
37 while(T--) {
38 scanf("%I64d%I64d%I64d%I64d%I64d",&n,&x,&y,&z,&t);
39 memset(f,0,sizeof(f));
40 now=0;
41 pre=1;
42 ll ans=x*t*n;
43 for(i=1; i<=n; i++) { ///第1到第i个不放红的
44 for(j=0; j<=i; j++) { ///第1个到第i个有j个green
45 k=i-j;///第1个到第i个有k个blue
46 if(k>0) {///第i个放蓝的
47 f[now][j][k]=f[pre][j][k-1] + j*y*z*(n-i);///放这个蓝的接下来n-i格每格增加了z时间
48 }
49 if(j>0) {///放绿的
50 f[now][j][k]=max(f[now][j][k],f[pre][j-1][k] + y*(t+k*z)*(n-i) );///放这个绿的每秒增加y伤害
51 }
52 ans=max(ans,f[now][j][k] + x*(n-i)*(t+k*z));
53 }
54 pre=now;
55 now^=1;
56 }
57 printf("Case #%I64d: %I64d\n",cas++,ans);
58 }
59 return 0;
60 }
hdu4939 Stupid Tower Defense （DP）,布布扣,bubuko.com