POJ-2940-Wine trading in Gergovia
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题大意为有n个等距的村庄,每个村庄要么买酒要么卖酒,把k个单位的酒从一个村庄运送到相邻的村庄需要k个单位劳动力,问最少需要多少劳动力才能满足所有村庄的需求。
由于最前面和最后面不相连,所以就比较简单了~
代码:
#include<cstdio>#include<cstring>#include<iostream>#include<cstdlib>using namespace std;const int maxn=10e4+10;int a[maxn],b[maxn];int main(){ int n; while(scanf("%d",&n)&&n) {int cou=0;for(int i=0;i<n;i++){ scanf("%d",&a[i]); if(a[i]>0)b[cou++]=i;}cou=0;long long ans=0;for(int i=0;i<n;i++){ if(a[i]<0) {while(a[i]<0){ int mini=min(-a[i],a[b[cou]]); a[i]+=mini; a[b[cou]]-=mini; ans+=mini*(abs(b[cou]-i)); if(!a[b[cou]])cou++;} }}printf("%I64d\n",ans); } return 0;}
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