Crixalis's Equipment

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

220 310 203 101 710 21 102 11

 

Sample Output

YesNo

#include <iomanip>#include <cstdio>#include <cstring>#include <vector>#include <iostream>#include <string.h>#include <algorithm>#include <set>#include <cmath>#define N 10005using namespace std;typedef long long ll;/*如果先放物品１，那么需要的空间是a1+b2；如果先放物品2，那么需要的空间是a2+b1；所以如果a1+b2<a2+b1的话，我们一定会选择先放物品１*/struct Equip {    int a;    int b;    bool operator < (const Equip &other) const {        return (a+other.b) < (b+other.a);    }};Equip equit[N];int T;int v, n;bool check() {    for (int i = 0; i < n; i++) {        if (v >= equit[i].b) v -= equit[i].a;        else return false;    }    return true;}int main(){    cin >> T;    while (T--) {        cin >> v >> n;        for (int i = 0; i < n; i++) {            cin >> equit[i].a >> equit[i].b;        }        sort(equit, equit+n);        if (check()) cout << "Yes" << endl;        else cout << "No" << endl;    }    return 0;}