hdu 1009 FatMouse' Trade (水题,贪心)
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小记:这题比较水
思路:将J[i]/F[i]的值进行从大到小排序,然后依次贪心
代码:
#include <iostream>#include <stdio.h>#include <string.h>#include <math.h>#include <stdlib.h>#include <map>#include <set>#include <vector>#include <stack>#include <queue>#include <algorithm>using namespace std;#define mst(a,b) memset(a,b,sizeof(a))#define REP(a,b,c) for(int a = b; a < c; ++a)#define eps 10e-8const int MAX_ = 1010;const int N = 100010;const int INF = 0x7fffffff;struct node{ int s, e; double dis;}t[MAX_];bool cmp(const node& a, const node& b){ if(a.dis < b.dis)return 0; return 1;}int main(){int T;int n, m;while(scanf("%d%d",&n, &m)) { if(n == -1 && m == -1)break; REP(i, 0, m) { scanf("%d%d", &t[i].s, &t[i].e); t[i].dis = t[i].s*1.0/t[i].e; } sort(t, t+m, cmp); double ans = 0; REP(i, 0, m){ if(t[i].e >= n){ ans += t[i].dis * n; break; } else { ans += t[i].s; n -= t[i].e; } } printf("%.3f\n", ans);}return 0;}
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