Greedy:Yogurt factory(POJ 2393)
题目大意:酸奶工厂每个星期都要制造酸奶,成本每单位x,然后每个星期要生产y,然后酸奶厂有个巨大的储存室,可以无限储存酸奶,而且酸奶的品质不会变坏,每天储存要每单位花费S,求最小的成本。
简直是初中生数学题,贪心法即可,假如当前酸奶成本是X,则如果储存酸奶,则下k个星期的(如果还使用这批酸奶),则成本为x+k*s,对比k个星期的成本就可以确定用哪个了。
水题
#include <iostream>
#include <functional>
#include <algorithm> typedef struct demand
{
int cost;
int units;
}Week; Week work[]; void Search(const int, const int); int main(void)
{
int N, S;
while (~scanf("%d%d", &N, &S))
{
for (int i = ; i < N; i++)
scanf("%d%d", &work[i].cost, &work[i].units);
Search(N, S);
}
return ;
} void Search(const int N, const int S)
{
int stop, now_cost, j;
long long ans = ; for (int pos = ; pos < N; )
{
now_cost = work[pos].cost;
for (stop = pos + , j = ; stop < N && now_cost + j*S < work[stop].cost; j++, stop++); for (int k = ; k < j; k++)//i是当前位置
ans += (long long)((now_cost + S*k) * work[pos + k].units);
pos = stop;
}
printf("%lld\n", ans);
}
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