Intel Code Challenge Final Round (Div. 1 + Div. 2, Combined) E. Goods transportation (非官方贪心解法)
2016-11-20 23:41
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题目链接:http://codeforces.com/contest/724/problem/E
题目大意:
有n个城市,每个城市有pi件商品,最多能出售si件商品,对于任意一队城市i,j,其中i<j,可以从城市i往j运输最多c件商品。 求最多一共能卖出多少件商品。 n<=10000
解法一(官方解法):
构造网络流,因为边太多,不可能直接跑最大流。 根据图的特殊性,考虑用dp求解最小割。
状态:dp[i,j]表示前i个中有j个和源点相通的最小割。
转移:如果第i个点不和源点相连,那么pi这条边一定要割掉,并且之前和源点相连的j个点,每个点会有一条边连向第i个点,这些边也要割掉。 花费是dp[i-1][j]+p[i]+j*c;
如果第i个点和源点相连,那么si这条边肯定要割掉。 花费是dp[i-1][j-1]+s[i];
故dp[i][j]=min(dp[i-1][j]+p[i]+j*c,dp[i-1][j-1]+s[i])。
最后答案就是min(dp
[0...n]) 时间复杂度O(n2)
代码:
View Code
题目大意:
有n个城市,每个城市有pi件商品,最多能出售si件商品,对于任意一队城市i,j,其中i<j,可以从城市i往j运输最多c件商品。 求最多一共能卖出多少件商品。 n<=10000
解法一(官方解法):
构造网络流,因为边太多,不可能直接跑最大流。 根据图的特殊性,考虑用dp求解最小割。
状态:dp[i,j]表示前i个中有j个和源点相通的最小割。
转移:如果第i个点不和源点相连,那么pi这条边一定要割掉,并且之前和源点相连的j个点,每个点会有一条边连向第i个点,这些边也要割掉。 花费是dp[i-1][j]+p[i]+j*c;
如果第i个点和源点相连,那么si这条边肯定要割掉。 花费是dp[i-1][j-1]+s[i];
故dp[i][j]=min(dp[i-1][j]+p[i]+j*c,dp[i-1][j-1]+s[i])。
最后答案就是min(dp
[0...n]) 时间复杂度O(n2)
代码:
#include <iostream> #include <cstdio> #include <cmath> #include <cstring> #include <algorithm> #include <vector> #include <map> #include <cstdlib> #include <set> using namespace std; #define X first #define Y second #define Mod 1000000007 #define N 10010 #define M 400010 typedef long long ll; typedef pair<int,int> pii; int n,c; int p ,s ; multiset<ll> st; int main() { //freopen("in.in","r",stdin); //freopen("out.out","w",stdout); ll ans,sum=0; scanf("%d%d",&n,&c); for (int i=1;i<=n;i++) scanf("%d",&p[i]),sum+=p[i]; for (int i=1;i<=n;i++) scanf("%d",&s[i]),st.insert(s[i]-p[i]+1ll*(n+1-i)*c); ans=sum; for (int i=1;i<=n;i++) { sum+= *st.begin()-1ll*i*c; st.erase(st.begin()); ans=min(ans,sum); } printf("%I64d\n",ans); return 0; }
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