3275: Number

Time Limit: 10 Sec  Memory Limit: 128 MB
Submit: 558  Solved: 246
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Output

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Source

我们能够发现仅仅有a和b奇偶性不同一时候才干满足题目条件。

然后对于全部奇数连边(s,i,a[i])，对于全部偶数连边(i,t,a[i])。对于满足题意的奇数i和偶数j连边(i,j,inf)。

最后跑一次最小割，从总和中减去最小割即为答案。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#define F(i,j,n) for(int i=j;i<=n;i++)
#define D(i,j,n) for(int i=j;i>=n;i--)
#define ll long long
#define pa pair<int,int>
#define maxn 3100
#define maxm 10000000
#define inf 1000000000
using namespace std;
struct edge_type
{int next,to,v;
}e[maxm];
int head[maxn],cur[maxn],dis[maxn],a[maxn];
int n,s,t,ans=0,tot=0,cnt=1;
inline int read()
{int x=0,f=1;char ch=getchar();while (ch<'0'||ch>'9'){if (ch=='-') f=-1;ch=getchar();}while (ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}return x*f;
}
inline void add_edge(int x,int y,int v)
{e[++cnt]=(edge_type){head[x],y,v};head[x]=cnt;e[++cnt]=(edge_type){head[y],x,0};head[y]=cnt;
}
inline bool bfs()
{queue<int>q;memset(dis,-1,sizeof(dis));dis[s]=0;q.push(s);while (!q.empty()){int tmp=q.front();q.pop();if (tmp==t) return true;for(int i=head[tmp];i;i=e[i].next) if (e[i].v&&dis[e[i].to]==-1){dis[e[i].to]=dis[tmp]+1;q.push(e[i].to);}}return false;
}
inline int dfs(int x,int f)
{if (x==t) return f;int tmp,sum=0;for(int &i=cur[x];i;i=e[i].next){int y=e[i].to;if (e[i].v&&dis[y]==dis[x]+1){tmp=dfs(y,min(f-sum,e[i].v));e[i].v-=tmp;e[i^1].v+=tmp;sum+=tmp;if (sum==f) return sum;}}if (!sum) dis[x]=-1;return sum;
}
inline void dinic()
{while (bfs()){F(i,1,t) cur[i]=head[i];ans+=dfs(s,inf);}
}
inline bool check(int x,int y)
{int tmp=x*x+y*y,rt=int(sqrt(tmp));if (rt*rt!=tmp) return false;if (x<y) swap(x,y);while (y) {tmp=x%y;x=y;y=tmp;}return (x==1);
}
int main()
{n=read();s=n+1;t=n+2;F(i,1,n){a[i]=read();tot+=a[i];if (a[i]&1) add_edge(s,i,a[i]);else add_edge(i,t,a[i]);}F(i,1,n) if (a[i]&1)F(j,1,n) if (!(a[j]&1))if (check(a[i],a[j])) add_edge(i,j,inf);dinic();printf("%d\n",tot-ans);
}