uva 6757 Cup of Cowards
Cup of Cowards (CoC) is a role playing game that has 5 different characters (Mage, Tank, Fighter,
Assassin and Marksman). A team consists of 5 players (one from each kind) and the goal is to kill a
monster with L life points. The monster dies if the total damage it gets is at least L. Each character
has a certain number of allowed hits, each hit has a certain damage and a certain cost (the cost and
damage might be different for each character). The team wants to kill the monster using the minimum
cost so they can perform better in later missions. They want your help to find the minimum cost they
will pay to kill the monster and the damage they should incur on it.
Input
Your program will be tested on one or more test cases. The first line of the input will be a single
integer T, the number of test cases (1 ≤ T ≤ 100). Followed by the test cases, the first line of each
test case contains 1 integer L (0 ≤ L ≤ 1012) representing the life points of the monster. Followed by
5 lines, each one contains 3 integers separated by a single space H D C representing the maximum
number of hits, the damage by each hit and the cost of each hit by one of the characters, respectively
(0 ≤ H ≤ 1, 000), (0 ≤ D, C ≤ 109
) and the sum of the maximum number of hits for all characters will
not be more than 1,000.
Output
For each test case, print a single line which contains 2 space separated integers, the first is the minimum
cost for the hits used to kill the monster and the second is the damage incurred upon the monster. If
there is more than one way to kill the monster using the same minimum cost, select the one with the
least damage and if there is no way to kill the monster print ‘We are doomed!!’ (without the quotes).
Sample Input
2
33
2 3 4
3 1 2
4 3 2
1 7 1
3 4 2
51
3 3 1
4 3 2
2 3 3
3 1 4
5 2 3
Sample Output
19 33
We are doomed!!

貌似中途相遇法,时间2500ms

view code#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
using namespace std;
typedef long long ll;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define iform "%I64d"
const int N = (1<<20)+10;
const ll INF = 1LL<<60;
ll _, L;
ll Min[N<<4],pos[N<<4], x[N<<1], xcnt;
ll d[100], c[100], cnt;
ll damage, cost;
struct node
{
ll v, c;
bool operator < (const node &o) const{
return v>o.v;
}
}one[N], two[N];
bool cmp(const node& a, const node &b)
{
return a.c<b.c;
}
void calc(node o[], int n, ll d[], ll c[], int &cnt)
{
cnt = 0;
for(int i=0; i<n; i++)
{
int k = cnt;
for(int j=0; j<k; j++)
{
if(o[j].v>=L) continue;
o[cnt].v = o[j].v+d[i];
o[cnt].c = o[j].c+c[i];
if(o[cnt].v>=L)
{
if(o[cnt].c<cost) cost = o[cnt].c, damage=o[cnt].v, cnt++;
else if(o[cnt].c==cost && damage>o[cnt].v) damage = o[cnt].v,cnt++;
}
else cnt++;
}
o[cnt].v = d[i];
o[cnt].c = c[i];
if(o[cnt].v>=L)
{
if(o[cnt].c<cost) cost = o[cnt].c, damage=o[cnt].v, cnt++;
else if(o[cnt].c==cost && damage>o[cnt].v) damage = o[cnt].v,cnt++;
}
else cnt++;
}
}
int ocnt, tcnt; void solve()
{
scanf(iform, &L);
ll num, dam, cos;
cnt = 0;
for(int i=1; i<=5; i++)
{
scanf(iform iform iform, &num, &dam, &cos);
ll k = 1;
while(num)
{
ll t = min(k, num);
d[cnt] = dam*t;
c[cnt] = cos*t;
cnt++;
num -= t;
k *= 2;
}
}
xcnt = 0;damage = -1, cost = INF;
ll n =cnt/2, m = cnt-n;
calc(one, n, d, c, ocnt);
calc(two, m, d+n, c+n, tcnt);
sort(one, one+ocnt, cmp);
sort(two, two+tcnt);
int i=0, j=0;
while(j<tcnt && two[j].v>=L) j++;
for(; i<ocnt&&j<tcnt; i++)
{
if(one[i].v>=L) continue;
else
{
while(j<tcnt && one[i].v+two[j].v>=L)
{
ll sumc = one[i].c+two[j].c;
ll sumv = one[i].v+two[j].v;
if(sumc<cost) damage=sumv,cost=sumc;
else if(sumc==cost) damage=sumv;
j++;
}
}
}
if(damage==-1) puts("We are doomed!!");
else cout<<cost<<""<<damage<<endl;
}
int main()
{
// freopen("in.txt", "r", stdin);
cin>>_;
while(_--) solve();
return 0;
}

uva 6757 Cup of Cowards(中途相遇法,貌似)的更多相关文章

  1. 紫书 例题8-3 UVa 1152(中途相遇法)

    这道题要逆向思维, 就是求出答案的一部分, 然后反过去去寻找答案存不存在. 其实很多其他题都用了这道题目的方法, 自己以前都没有发现, 这道题专门考这个方法.这个方法可以没有一直往下求, 可以省去很多 ...

  2. 紫书 习题 8-16 UVa 1618 (中途相遇法)

    暴力n的四次方, 然而可以用中途相遇法的思想, 分左边两个数和右边两个数来判断, 最后合起来判断. 一边是n平方logn, 合起来是n平方logn(枚举n平方, 二分logn) (1)两种比较方式是相 ...

  3. 【uva 1152】4 Values Whose Sum is Zero(算法效率--中途相遇法+Hash或STL库)

    题意:给定4个N元素几个A,B,C,D,要求分别从中选取一个元素a,b,c,d使得a+b+c+d=0.问有多少种选法.(N≤4000,D≤2^28) 解法:首先我们从最直接最暴力的方法开始思考:四重循 ...

  4. LA 2965 Jurassic Remains (中途相遇法)

    Jurassic Remains Paleontologists in Siberia have recently found a number of fragments of Jurassic pe ...

  5. HDU 5936 Difference 【中途相遇法】(2016年中国大学生程序设计竞赛(杭州))

    Difference Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total ...

  6. 【中途相遇法】【STL】BAPC2014 K Key to Knowledge (Codeforces GYM 100526)

    题目链接: http://codeforces.com/gym/100526 http://acm.hunnu.edu.cn/online/?action=problem&type=show& ...

  7. 高效算法——J 中途相遇法,求和

    ---恢复内容开始--- J - 中途相遇法 Time Limit:9000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Su ...

  8. 【UVALive】2965 Jurassic Remains(中途相遇法)

    题目 传送门:QWQ 分析 太喵了~~~~~ 还有中途相遇法这种东西的. 嗯 以后可以优化一些暴力 详情左转蓝书P58 (但可能我OI生涯中都遇不到正解是这个的题把...... 代码 #include ...

  9. uva1152 - 4 Values whose Sum is 0(枚举,中途相遇法)

    用中途相遇法的思想来解题.分别枚举两边,和直接暴力枚举四个数组比可以降低时间复杂度. 这里用到一个很实用的技巧: 求长度为n的有序数组a中的数k的个数num? num=upper_bound(a,a+ ...

随机推荐

  1. 我们的相识,总是那么巧。-------eclipse中搭建maven项目

    一.我们就来谈下eclipse中搭建maven web工程的步骤!虽然就是一个简单的例子,但是过程是很艰辛的. 首先我们看一下eclipse的封面,下面就是刚打开的华丽封面哦 其次我安装了eclips ...

  2. ssh默认端口更改后,如何正常使用git?

    由于安全或者其它原因,我们可能会修改默认的SSH服务端口号,默认情况下,已有的git项目在pull或者push的时候会报错. 现在假设原来的项目的remote设置为git@domain.com:Pro ...

  3. jQuery cbpContentSlider 滑动切换

    cbpContentSlider是一款选项卡插件,只要按照以下html结构就可以自动生成菜单切换内容特效. 在线实例 实例演示 使用方法 <div id="cbp-contentsli ...

  4. 用Microsoft.Solver.Foundation进行线性规划,为WPF应用添加智能

    在管理信息系统的开发过程中,往往会涉及到一些线性规划数学模型,例如资源配置优化.微软的Microsoft.Solver.Foundation是一个数学库,可以很好的对线性规划问题进行求解.关于它的细节 ...

  5. Typecast 免费了!献给设计师们的礼物

    TypeCast 让你可以从 Fonts.com.TypeKit.FontDeck 和 Google 这些字体供应和商选择字体,而且能非常方便的比较这些字体使用效果.如果你想获得用户对这些字体效果的反 ...

  6. 【经验之谈】前端面试知识点总结(HTML相关)——附答案

    目录 一.HTML部分 1.浏览器页面有哪三层构成,分别是什么,作用是什么? 2.HTML5的优点与缺点? 3.Doctype作用? 严格模式与混杂模式如何区分?它们有何意义? 4.HTML5有哪些新 ...

  7. 【NodeJS 学习笔记02】入门资源很重要

    前言 在我映像中,异步最早出现与ajax,当时我还在搞.net,然后.net居然出了一个异步的控件...... 虽然我最后知道了他不是异步的......然后,前端异步用得特别多,如果不是异步的程序,你 ...

  8. 自定义easyui整数或者数字、字母或者下划线验证方法

    $.extend($.fn.validatebox.defaults.rules, { intOrFloat: {// 验证整数或小数 validator: function (value) { re ...

  9. 【移动适配】移动Web怎么做屏幕适配(三)

    复杂纷扰的世界背后,总会有万变不离其宗的简单规则 啃先生 Mar.8th.2016 壹 | Fisrt 前面写了两篇移动适配相关的文章: <移动Web怎么做屏幕适配(一)>重点介绍了怎样利 ...

  10. JavaScript学习10 JS数据类型、强制类型转换和对象属性

    JavaScript学习10 JS数据类型.强制类型转换和对象属性 JavaScript数据类型 JavaScript中有五种原始数据类型:Undefined.Null.Boolean.Number以 ...