Hdoj 3177 crixment & #39; s Equipment [greedy], hdojcrixment

Hdoj 3177 Crixalis's Equipment [greedy], hdojcrix.pdf

Crixalis's EquipmentTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission (s): 3073 Accepted Submission (s): 1250

Problem descriptioncrix‑sand King used to be a giant scorpion (scorpion) 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 crixincludecides 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 crix?has N equipment, each of them needs Ai units of space. when dragging his equipment into the hole, crix?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 crix?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. crix?wonders if he can move all his equipment into the new hole and he turns to you for help.
 
InputThe 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.
 
OutputFor each case output "Yes" if crix?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

Sort (B-a) by the difference .....

Code:

#include <stdio.h>#include <string.h>#include <algorithm>#define M 1005using std::sort;struct node{    int a, b;}s[M];int cmp(node a, node b){    return (a.b-a.a)>(b.b-b.a);}int main(){    int t, v, n;    scanf("%d", &t);    while(t --){        scanf("%d%d", &v, &n);        int i;        for(i = 0; i < n; i ++) scanf("%d%d", &s[i].a, &s[i].b);        sort(s, s+n, cmp);        for(i = 0; i < n; i ++){            if(v>=s[i].b){                v-=s[i].a;            }            else break;        }        if(i == n) printf("Yes\n");        else printf("No\n");    }     return 0;}