Crixalis's Equipment Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2562 Accepted Submission(s): 1056
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.
0<T<= 10, 0<V<10000, 0<N<1000, 0 <Ai< V, Ai <= Bi < 1000.
<span style="font-size:18px;">#include<iostream> #include<cstdio> #include<algorithm> using namespace std; struct Node { int a,b; int c; }s[1005]; bool cmp(Node a,Node b) { return a.c>b.c; } int main() { int T,V,N,i; scanf("%d",&T); while(T--) { scanf("%d%d",&V,&N); for(i=0;i<N;i++) { scanf("%d%d",&s[i].a,&s[i].b); s[i].c=s[i].b-s[i].a; } sort(s,s+N,cmp); for(i=0;i<N;i++) { if(s[i].b>V) { printf("No\n"); break; } else V=V-s[i].a; } if(i==N) printf("Yes\n"); } return 0; } </span>