Crixalis's Equipment

Problem H: Crixalis's Equipment

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 73  Solved: 25
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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

220 310 203 101 710 21 102 11

Sample Output

YesNo

HINT

第一眼看过去，以为直接按bi的大小来贪就行了，结果wa ，后来稍微举一下例子就能懂了

eg: 两个物品 (1,5),(4,6),如果先装第一个后装第二个所需要的空间为（max(5,1+6)==7），反之为（max(6,4+5)==9），可以看出坑定先选第一个再选第二个能装的更多，不难理解，差值越大的时候，证明我用较小的空间存储到了较大的装备，所肯定优先选择差值越大的啦

#include<cstdio>#include<cstring>#include<cctype>#include<algorithm>#include<set>#include<cstring>#include<string>#include<iostream>#include<cmath>#include<map>#include<vector>#include<stack>using namespace std;struct equi{    int ai;    int bi;}e[1005];bool cmp(equi a,equi b){    return a.bi-a.ai>b.bi-b.ai;}int main(){    int t;    scanf("%d",&t);    while(t--){            memset(e,0,sizeof(e));        int v,n;        scanf("%d %d",&v,&n);        for(int i=0;i<n;i++)        scanf("%d %d",&e[i].ai,&e[i].bi);        sort(e,e+n,cmp);        int flag=0;        for(int i=0;i<n;i++){            if(v>=e[i].bi){                v-=e[i].ai;            }            else {                flag=1;                break;            }        }        if(!flag)            printf("Yes\n");        else            printf("No\n");    }    return 0;}