Problem C: The Dragon of Loowater
Once upon a time, in the Kingdom of Loowater, a minor nuisance turnedinto a major problem.
The shores of Rellau Creek in central Loowater had always been a primebreeding ground for geese. Due to the lack of predators, the geesepopulation was out of control. The people of Loowater mostly kept clearof the geese. Occasionally, a goose would attack
one of the people,and perhaps bite off a finger or two, but in general, the peopletolerated the geese as a minor nuisance.
One day, a freak mutation occurred, and one of the geese spawned amulti-headed fire-breathing dragon. When the dragon grew up, he threatenedto burn the Kingdom of Loowater to a crisp. Loowater had a major problem.The king was alarmed, andcalled on his knights
to slay the dragon and save the kingdom.
The knights explained: "To slay the dragon, we must chop off all itsheads. Each knight can chop off one of the dragon's heads. The headsof the dragon are of different sizes. In order to chop off a head,a knight must be at least as tall as the diameter of
the head.The knights' union demands that for chopping off a head, a knightmust be paid a wage equal to one gold coin for each centimetreof the knight's height."
Would there be enough knights to defeat the dragon? The king called onhis advisors to help him decide how many and which knights to hire.After having lost a lot of money building Mir Park, the king wantedto minimize the expense of slaying the dragon. As
one of the advisors,your job was to help the king. You took it very seriously: if you failed,you and the whole kingdom would be burnt to a crisp!
Input Specification:
The input contains several test cases. The first line of each test casecontains two integers between 1 and 20000 inclusive, indicating thenumbern of heads that the dragon has, and the number
mof knights in the kingdom. The nextn lines each contain aninteger, and give the diameters of the dragon's heads, in centimetres.The followingm lines each contain an integer, and specifythe heights of the knights of Loowater, also
in centimetres.
The last test case is followed by a line containing:
0 0
Output Specification:
For each test case, output a line containing the minimum number ofgold coins that the king needs to pay to slay the dragon. If it is notpossible for the knights of Loowater to slay the dragon, output theline:
Loowater is doomed!
Sample Input:
2 3 5 4 7 8 4 2 1 5 5 10 0 0
Output for Sample Input:
11
Loowater is doomed!
题目讲故事废话讲了很多
主要是: 有一个由鸭子变异的龙, 有N个头, 每个头半径不同
还有M个骑士, 骑士有不同的身高
国王想灭掉这头龙, 就要砍掉龙的所有头
一个骑士只能砍一个头, 而且每个骑士只能砍半径小于等于自身身高的龙头
如果能杀死龙, 那国王给骑士赏钱, 和身高1比1的钱, 例如某个骑士身高是7 那他就拿7的酬劳
要求输出国王最低要赏多少钱
如果不能杀死, 那就输出 Loowater is doomed!
做法比较简单, 两次生序排序再循环一下就OK了~
AC代码:
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<algorithm>
using namespace std;
int cmp(const void *a, const void *b) {
return *(int *)a - *(int *)b;
}
int n, m;
int head[25000];
int kni[25000];
int sum;
int mark;
int main() {
while(scanf("%d%d", &n, &m) != EOF) {
if(n == 0 && m == 0)
break;
sum = 0;
mark = 1;
for(int i = 0; i < n; i++)
scanf("%d", &head[i]);
for(int i = 0; i < m; i++)
scanf("%d", &kni[i]);
if(n > m)
mark = 0;
sort(head, head+n);
sort(kni, kni+m);
int count = 0;
for(int i = 0; i < n; i++) {
for(int j = 0; j < m; j++) {
if(kni[j] >= head[i]) {
sum += kni[j];
kni[j] = -1;
count++;
break;
}
}
}
if(count == n && mark == 1)
printf("%d\n", sum);
else
printf("Loowater is doomed!\n");
}
return 0;
}