Codeforces (441B) Round #252 (Div. 2) B题
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题目链接:http://codeforces.com/contest/441/problem/B
题目大意:有一个果园里有很多树,上面有很多果实,为了不然成熟的果实腐烂,必须在两天之内收集起来。给出果园有的树,以及该树上的果实个数,工人每天可以采集的上限,求出这段时间之后,能收集到的最大值
很简单,维护一个一维数组ans[],首先将数据存在结构体中,再按果实成熟的日期ai升序排序,优先将果实计入ans[ai]中,多的放入ans[ai+1]里面,如果大于上限,则记为上限大小,最后这个数组全部元素的和就是答案了。
#include <iostream>#include <cstdio>#include <algorithm>#define MAX_N 3000using namespace std;struct fruit{ int x,y;};bool cmp(fruit a,fruit b){ return a.x<b.x;}fruit a[MAX_N+1];int ans[MAX_N+1];int main(){ int n,v,maxi=0; scanf("%d%d",&n,&v); for(int i=0;i<n;i++) { scanf("%d%d",&a[i].x,&a[i].y); maxi=max(maxi,a[i].x); } sort(a,a+n,cmp); for(int i=0;i<n;i++) { if(ans[a[i].x-1]+a[i].y<=v) { ans[a[i].x-1]+=a[i].y; } else { if(ans[a[i].x-1]+ans[a[i].x]+a[i].y-v>v) ans[a[i].x]=v; else ans[a[i].x]+=ans[a[i].x-1]+a[i].y-v; ans[a[i].x-1]=v; } } int Mans=0; for(int i=0;i<=maxi;i++) Mans+=ans[i]; printf("%d\n",Mans); return 0;}
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