2014
03-06

# Crixalis’s Equipment

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.

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.

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.

2

20 3
10 20
3 10
1 7

10 2
1 10
2 11

Yes
No

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

停放体积         移动体积

第一件物品          a1             b1

第二件物品          a2             b2

a2+b1为先放第二件物品 后放第一件物品的最大瞬时体积

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

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

#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");
}

}
}

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