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Crixalis's Equipment
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4590 Accepted Submission(s): 1875
Problem Description Crixalis - Sand King used to be a giant scorpion(蝎子) in the deserts of Kalimdor. Though he's a guardian of Lich King now, he keeps the living habit of a scorpion like living underground and digging holes.
Someday Crixalis decides to move to another nice place and build a new house for himself (Actually it's just a new hole). As he collected a lot of equipment, he needs to dig a hole beside his new house to store them. This hole has a volume of V units, and Crixalis has N equipment, each of them needs Ai units of space. When dragging his equipment into the hole, Crixalis finds that he needs more space to ensure everything is placed well. Actually, the ith equipment needs Bi units of space during the moving. More precisely Crixalis can not move equipment into the hole unless there are Bi units of space left. After it moved in, the volume of the hole will decrease by Ai. Crixalis wonders if he can move all his equipment into the new hole and he turns to you for help.
Input The first line contains an integer T, indicating the number of test cases. Then follows T cases, each one contains N + 1 lines. The first line contains 2 integers: V, volume of a hole and N, number of equipment respectively. The next N lines contain N pairs of integers: Ai and Bi.
0<T<= 10, 0<V<10000, 0<N<1000, 0 <Ai< V, Ai <= Bi < 1000.
Output For each case output "Yes" if Crixalis can move all his equipment into the new hole or else output "No".
Sample Input
2 20 3 10 20 3 10 1 7 10 2 1 10 2 11
Sample Output
Yes No
题目大意:Crixalis要换到新房子里了,他有n个设备要搬到洞里,这个洞的大小为v个单元,移动设备时需要占用b个单元大小,搬到洞里后,自身占a个体积,如果设备可以全部搬到洞里,输出“Yes”,否则输出“No”。
思路:贪心算法,每次搬运使剩下的空间最大。
假设有两个设备
A: a1 b1
B: a2 b2
先搬A,瞬间占用空间最大为 a1+b2;先搬B,瞬间占用空间最大为a2+b1.为了使剩下的体积最大,我们要选择这两个所占体积最小的,
假如先选A,则需满足 a1+b2 < a2+b1 得出 b1-a1 > b2-a2
所以我们应该选择搬运设备与自身体积所占空间的差最大的来搬运
#include<stdio.h>
#include<iostream>
#include<algorithm>
using namespace std;
struct space{
int a, b, c;
}room[1010];
int cmp(space x, space y){
return x.c > y.c;
}
int main(){
int t, v, n;
scanf("%d", &t);
while(t--){
scanf("%d%d", &v, &n);
for(int i = 0; i < n; i++){
scanf("%d%d", &room[i].a, &room[i].b);
room[i].c = room[i].b-room[i].a; //搬运设备与自身体积的差
}
sort(room, room+n, cmp);
int flag = 1;
for(int i = 0; i < n; i++){
if(room[i].b > v){
flag = 0;
break;
}else{
v -= room[i].a;
}
}
if(flag){
printf("Yes\n");
}else
printf("No\n");
}
return 0;
}
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