首页 > ACM题库 > HDU-杭电 > HDU 3177-Crixalis’s Equipment-贪心-[解题报告]HOJ
2014
03-06

HDU 3177-Crixalis’s Equipment-贪心-[解题报告]HOJ

Crixalis’s Equipment

问题描述 :

Broken PixelCrixalis – 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.

输入:

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.

输出:

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.

样例输入:

2

20 3
10 20
3 10
1 7

10 2
1 10
2 11

样例输出:

Yes
No

题目大意:向体积为v的山洞中搬运n个物品,每个物品具有(a,b) 属性。其中a是停放体积,b是移动体积。输出这个山东是否能放下这n个物品

解题思路:

1)当前物品能否放进山洞取决于当前物品的的移动体积是否小于山洞当前的剩余体积。

2)对这些物品进行排序 按照顺序依次进入洞中 排序要尽可能使得所有的东西都能进入洞中

这是一个贪心的问题  

                                   停放体积         移动体积

                  第一件物品          a1             b1

  

                  第二件物品          a2             b2

假设这两件物品的移动体积都不大于洞的体积V

那么将单独比较两个物品的时候会发现  a1+b2为先放第一件物品 后放第二件物品的最大瞬时体积

                                                  a2+b1为先放第二件物品 后放第一件物品的最大瞬时体积

我们应该选择a1+b2和a2+b1中比较小的先放

那么从2件物品 扩展到N件物品 假设n件物品的移动体积都不大于洞的体积V(如果有大于的 那么结果必然是NO)

将N件物品按照a1+b2<a2+b1进行排序 然后依次放入洞中

3)也就是说,每次向山洞中放移动体积和停放体积最大的那个物品

代码如下:

/*
 * 3177_2.cpp
 *
 *  Created on: 2013年8月10日
 *      Author: Administrator
 */

#include <iostream>
using namespace std;
struct Node{
	int f;
	int l;
};

bool compare(Node a , Node b){
	return a.f + b.l < b.f + a.l;
}


int main(){
	int t;
	scanf("%d",&t);

	while(t--){
		int v , n;
		scanf("%d%d",&v,&n);
		int i;
		Node nodes[n];
		for(i = 0 ; i < n ; ++i){
			scanf("%d%d",&nodes[i].f,&nodes[i].l);
		}

		sort(nodes,nodes+n,compare);

		bool flag = true;
		for(i = 0 ; i < n ; ++i){
			if(nodes[i].l > v){
				printf("No\n");
				flag = false;
				break;
			}else{
				v -= nodes[i].f;
			}
		}


		if(flag){
			printf("Yes\n");
		}

	}
}

参考:http://blog.csdn.net/hjd_love_zzt/article/details/9877115


  1. 算法是程序的灵魂,算法分简单和复杂,如果不搞大数据类,程序员了解一下简单点的算法也是可以的,但是会算法的一定要会编程才行,程序员不一定要会算法,利于自己项目需要的可以简单了解。

  2. #!/usr/bin/env python
    def cou(n):
    arr =
    i = 1
    while(i<n):
    arr.append(arr[i-1]+selfcount(i))
    i+=1
    return arr[n-1]

    def selfcount(n):
    count = 0
    while(n):
    if n%10 == 1:
    count += 1
    n /= 10
    return count