Problem C: The Dragon of Loowater Once upon a time, in the Kingdom of Loowater, a minor nuisance turned into a major problem. The shores of Rellau Creek in central Loowater had always been a prime breeding ground for geese. Due to the lack of predato…
Problem C: The Dragon of Loowater Once upon a time, in the Kingdom of Loowater, a minor nuisance turned into a major problem. The shores of Rellau Creek in central Loowater had always been a prime breeding ground for geese. Due to the lack of predato…
Once upon a time, in the Kingdom of Loowater, a minor nuisance turned into a major problem. The shores of Rellau Creek in central Loowater had always been a prime breeding ground for geese. Due to the lack of predators, the geese population was out o…
Once upon a time, in the Kingdom of Loowater, a minor nuisance turned into a major problem.The shores of Rellau Creek in central Loowater had always been a prime breeding ground for geese.Due to the lack of predators,the geese population was out of c…
简单贪心 龙头的直径和人的佣金排序,价值小的人和直径小的配 #include<iostream> #include<cstdio> #include<cmath> #include<cstring> #include<algorithm> typedef long long ll; #define N 20005 using namespace std; int a[N],b[N]; int main(){ int m,n; int i,j; w…
贪心策略:一个直径为X的头颅,应该让雇佣费用满足大于等于X且最小的骑士来砍掉,这样才能使得花费最少. AC代码 #include <cstdio> #include <cmath> #include <algorithm> #include <cstring> #include <utility> #include <string> #include <iostream> #include <map> #inc…
题意:有n个条龙,在雇佣勇士去杀,每个勇士能力值为x,只能杀死头的直径y小于或等于自己能力值的龙,只能被雇佣一次,并且你要给x赏金,求最少的赏金. 析:很简单么,很明显,能力值高的杀直径大的,低的杀直径小的.所以我们先对勇士能力值从小到大排序,然后对龙的直径从小到大排序, 然后扫一遍即可,如某个勇士杀不龙,就可以跳过,扫到最后,如果杀完了就结束,输出费用,否则就是杀不完. 代码如下: #include <iostream> #include <cstdio> #include &l…
Problem C: The Dragon of Loowater Once upon a time, in the Kingdom of Loowater, a minor nuisance turned into a major problem. The shores of Rellau Creek in central Loowater had always been a prime breeding ground for geese. Due to the lack of predato…
你的王国里有一条n个头的恶龙,你希望雇佣一些骑士把它杀死(也就是砍掉所有的头).村里有m个骑士可以雇佣,一个能力值为 x 的骑士可以砍掉恶龙一个直径不超过 x 的头,且需要支付 x 个金币.如何雇佣骑士才能砍掉恶龙所有的头,并且支付最小的金币?注意,一个骑士只能砍一个头并且仅能被雇佣1次 因为要保证用的钱最少,所以先把骑士按照能力值从小到大进行排序.然后从最小的开始一个一个进行匹配.在进行匹配的时候又出现一个问题,那就是每个骑士只能雇佣一次.这里有2个处理方法,第一个是开一个数组用来标记该骑士是…
---恢复内容开始--- 题目: Once upon a time, in the Kingdom of Loowater, a minor nuisance turned into a major problem. The shores of Rellau Creek in central Loowater had always been a prime breeding ground for geese. Due to the lack of predators, the geese pop…
题目传送门 /* 题意:n个头,m个士兵,问能否砍掉n个头 贪心/思维题:两个数组升序排序,用最弱的士兵砍掉当前的头 */ #include <cstdio> #include <cstring> #include <algorithm> using namespace std; ; const int INF = 0x3f3f3f3f; int a[MAXN], b[MAXN]; int main(void) //UVA 11292 The Dragon of Loo…
1405. 中古世界的恶龙[The Drangon of Loowater,UVa 11292] ★   输入文件:DragonUVa.in   输出文件:DragonUVa.out   简单对比时间限制:1 s   内存限制:256 MB [题目描述] King(CH)的王国里有一条n个头的恶龙,他希望雇佣一些骑士把它杀死(即砍掉所有的头).小酒馆里有m个骑士可以雇佣,一个能力值为x的骑士可以砍掉一个直径不超过x的头,且需要支付x个金币.如何雇用骑士才能砍掉所有恶龙的头,且需要支付的金币最少?…
***从今天开始自学算法. ***代码是用c++,所以顺便再自学一下c++ 例题1  勇者斗恶龙(The Dragon of Loowater, UVa 11292) 你的王国里有一条n个头的恶龙,你希望雇一些骑士把它杀死(即砍掉所有头).村里有m个骑士可以雇佣,一个能力值为x的骑士可以砍掉恶龙一个直径不超过x的头,且需要支付x个金币.如何雇佣骑士才能砍掉恶龙的所有头,且需要支付的金币最少?注意,一个骑士只能砍一个头(且不能被雇佣两次). [输入格式] 输入包含多组数据.每组数据的第一行为正整数…
  The Dragon of Loowater Once upon a time, in the Kingdom of Loowater, a minor nuisance turned into a major problem. The shores of Rellau Creek in central Loowater had always been a prime breeding ground for geese. Due to the lack of predators, the g…
Problem C: The Dragon of Loowater Once upon a time, in the Kingdom of Loowater, a minor nuisance turned into a major problem. The shores of Rellau Creek in central Loowater had always been a prime breeding ground for geese. Due to the lack of predato…
option=com_onlinejudge&Itemid=8&page=show_problem&problem=2267" style="color:blue; text-decoration:none">Dragon of Loowater Time Limit: 1000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu [Submit]   [Go Back]  …
首先先看一下这道题的英文原版... 好吧,没看懂... 大体意思就是: 有一条n个头的恶龙,现在有m个骑士可以雇佣去杀死他,一个能力值为x的勇士可以砍掉直径不超过x的头,而且需要支付x个金币.如何雇佣才能砍掉所有的头且支付最少的金币,注意一个勇士只能砍一个头,也只能被雇佣一次. 输入包含多组数据,每组数据第一行为正整数n,m(1<=n,m<=20000),以下n行每行为一个整数,就是恶龙的头的直径,以下m行每行为一个整数,就是每个骑士的能力值.输入结束标志位m=n=0. 输出格式对于每组数据输…
http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=2267 题目大意:n条恶龙,m个勇士,用勇士来杀恶龙.一个勇士只能杀一个恶龙.而且勇士只能杀直径不超过自己能力值的恶龙.每个勇士需要支付能力值一样的金币.问杀掉所有恶龙需要的最少金币? 题意分析:先排序,再进行筛选. #include<stdio.h>…
题目大意:   你的王国里有一条n个头的恶龙,你希望雇一些骑士把它杀死(即砍掉所有头).村里有m个骑士可以雇佣,一个能力值为x的骑士可以砍掉恶龙一个直径不超过x的头,且需要支付x个金币.如何雇佣骑士才能砍掉恶龙的所有头,且需要支付的金币最少?注意,一个骑士只能砍一个头(且不能被雇佣两次).  输入格式 输入包含多组数据.每组数据的第一行为正整数n和m(1≤n,m≤20 000):以下n行每行为一个整数,即恶龙每个头的直径:以下m行每行为一个整数,即每个骑士的能力.输入结束标志为n=m=0. 输出…
题意:有一条有n个头的恶龙,有m个骑士去砍掉它们的头,每个骑士可以砍直径不超过x的头,问怎样雇佣骑士,使花的钱最少 把头的直径从小到大排序,骑士的能力值也从小到大排序,再一个一个地去砍头 #include<iostream> #include<cstdio> #include<cstring> #include <cmath> #include<stack> #include<vector> #include<map> #…
https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=2267 有一条n个头的恶龙,现在有m个骑士可以雇佣去杀死他,一个能力值为x的勇士可以砍掉直径不超过x的头,而且需要支付x个金币.如何雇佣才能砍掉所有的头且支付最少的金币,注意一个勇士只能砍一个头,也只能被雇佣一次. 输入包含多组数据,每组数据第一行为正整数n,…
Once upon a time, in the Kingdom of Loowater, a minor nuisance turned into a major problem. The shores of Rellau Creek in central Loowater had always been a prime breeding ground for geese. Due to the lack of predators, the geese population was out o…
思路:先将龙和士兵进行分别排序从小到大.然后,每次找当前最小龙的第一个大于它的骑手之后退出,开始下一个龙,重复上一次操作. #include<iostream> #include<algorithm> using namespace std; ; int a[maxn], b[maxn]; int n, m; int main(){ while(cin>>n>>m){ )break; ;i<=n;++i)cin>>a[i]; ;i<=…
https://vjudge.net/problem/UVA-11292 题意:有n条任意个头的恶龙,你希望雇一些其实把它杀死.一个能力值为x的骑士可以砍掉恶龙一个直径不超过x的头,且需要支付x个金币.输出最少金币. 思路: 对两者进行排序.依次比较,头少的龙尽量用能力值小的骑士去砍. #include<iostream> #include<cstring> #include<string> #include<algorithm> #include<v…
        Time Limit: 1sec    Memory Limit:32MB  Description Once upon a time, in the Kingdom of Loowater, a minor nuisance turned into a major problem. The shores of Rellau Creek in central Loowater had always been a prime breeding ground for geese. D…
Once upon a time, in the Kingdom of Loowater, a minor nuisance turned into a major problem. The shores of Rellau Creek in central Loowater had always been a prime breeding ground for geese. Due to the lack of predators, the geese population was out o…
水题,排序遍历即可 #include<iostream> #include<cstdio> #include<algorithm> using namespace std; ; int drag[maxn], kn[maxn]; int main() { int n, m, cost; while (cin >> n >> m) { && m == )break; ;i < n;i++)cin >> drag[i…
蓝书P1, 很简单的一个贪心选择,用能力小的去砍小的.本来想双重循环,哎,傻逼了,直接遍历选手,碰到能砍的就砍掉. #include <stdio.h> #include <algorithm> using namespace std; #define MAXN 20005 int n,m; int nn[MAXN],mm[MAXN]; int main() { freopen("input.txt","r",stdin); while(sc…
题目大意: 你的王国里有一条n个头的恶龙,你希望雇佣一些骑士把它杀死(即砍掉所有头).村里有m个骑士可以雇佣,一个能力值为x的骑士可以砍掉恶龙一个直径不超过x的头,且需要支付x个金币.如何雇佣骑士才能砍掉龙的所有头,且需要支付的金币最少?注意,一个骑士只能砍一个头.(且不能被雇佣两次). 输入格式 输入包含多组数据.每组数据的第一行为正整数n和m(1<=n,m<=20000):以下n行每行为一个整数,即恶龙每个头的直径:以下m行每行为一个整数,即每个骑士的能力.输入结束标志为n=m=0. 输出…
#include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<string> #include<cmath> #include<map> #include<set> #include<vector> #include<algorithm> #include<stack> #in…