【BZOJ-3275&3158】Number&千钧一发最小割

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3158: 千钧一发

Time Limit: 10 Sec Memory Limit: 512 MB
Submit: 984 Solved: 359
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Description

技术分享

Input

第一行一个正整数N。

第二行共包括N个正整数，第个正整数表示Ai。

第三行共包括N个正整数，第个正整数表示Bi。

Output

共一行，包括一个正整数，表示在合法的选择条件下，可以获得的能量值总和的最大值。

Sample Input

4
3 4 5 12
9 8 30 9

Sample Output

HINT

1<=N<=1000,1<=Ai,Bi<=10^6

Source

Katharon+#1

Solution

一开始没仔细读题，看成同时满足两个，WA了几波后才发现看错了...

对奇数偶数讨论一下,奇数之间肯定满足1.偶数之间肯定满足2

最小割就好

Code

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<queue>
using namespace std;
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;
}
int Gcd(int a,int b)
{
    if (b==0) return a; else return Gcd(b,a%b);
}
bool check(long long a,long long b)
{
    if (Gcd(a,b)>1) return true;
    long long T=sqrt(a*a+b*b); if (T*T!=a*a+b*b) return true;
    return false; 
}
#define maxn 5000
#define maxm 1000010
int N,A[maxn],B[maxn],tot;
struct Edgenode{int next,to,cap;}edge[maxm];
int head[maxn],cnt=1;
void add(int u,int v,int w) 
{
    cnt++;
    edge[cnt].next=head[u]; head[u]=cnt; edge[cnt].to=v; edge[cnt].cap=w;
}
void insert(int u,int v,int w) {add(u,v,w); add(v,u,0);}
int cur[maxn],dis[maxn],S,T;
bool bfs()
{
    queue<int> q;
    memset(dis,-1,sizeof(dis));
    q.push(S); dis[S]=0;
    while (!q.empty())
        {
            int now=q.front(); q.pop();
            for (int i=head[now]; i; i=edge[i].next)
                if (edge[i].cap && dis[edge[i].to]==-1)
                    dis[edge[i].to]=dis[now]+1,q.push(edge[i].to);
        }
    return dis[T]!=-1;
}
int dfs(int loc,int low)
{
    if (loc==T) return low;
    int w,used=0;
    for (int i=cur[loc]; i; i=edge[i].next)
        if (edge[i].cap && dis[edge[i].to]==dis[loc]+1)
            {
                w=dfs(edge[i].to,min(low-used,edge[i].cap));
                edge[i].cap-=w; edge[i^1].cap+=w; used+=w;
                if (edge[i].cap) cur[loc]=i; if (used==low) return low;
            }
    if (!used) dis[loc]=-1;
    return used;
}
#define inf 0x7fffffff
int dinic()
{
    int tmp=0;
    while (bfs())
        {
            for (int i=S; i<=T; i++) cur[i]=head[i];
            tmp+=dfs(S,inf);
        }
    return tmp;
}
void Build()
{
    S=0,T=N+1;
    for (int i=1; i<=N; i++) 
        if ((A[i]%2)) insert(S,i,A[i]); else insert(i,T,A[i]);
    for (int i=1; i<=N; i++)
        for (int j=1; j<=N; j++)    
            if ((A[i]%2) && !(A[j]%2) && !check(A[i],A[j]))
                insert(i,j,inf);
}
int main()
{
    N=read();
    for (int i=1; i<=N; i++) A[i]=read(),tot+=A[i];
    //for (int i=1; i<=N; i++) B[i]=read(),tot+=B[i];
    Build();
    int maxflow=dinic();
    printf("%d\n",tot-maxflow);
    return 0;
}

BZOJ-3275

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<queue>
using namespace std;
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;
}
int Gcd(int a,int b)
{
    if (b==0) return a; else return Gcd(b,a%b);
}
bool check(long long a,long long b)
{
    if (Gcd(a,b)>1) return true;
    long long T=sqrt(a*a+b*b); if (T*T!=a*a+b*b) return true;
    return false; 
}
#define maxn 3000
#define maxm 1000010
int N,A[maxn],B[maxn],tot;
struct Edgenode{int next,to,cap;}edge[maxm];
int head[maxn],cnt=1;
void add(int u,int v,int w) 
{
    cnt++;
    edge[cnt].next=head[u]; head[u]=cnt; edge[cnt].to=v; edge[cnt].cap=w;
}
void insert(int u,int v,int w) {add(u,v,w); add(v,u,0);}
int cur[maxn],dis[maxn],S,T;
bool bfs()
{
    queue<int> q;
    memset(dis,-1,sizeof(dis));
    q.push(S); dis[S]=0;
    while (!q.empty())
        {
            int now=q.front(); q.pop();
            for (int i=head[now]; i; i=edge[i].next)
                if (edge[i].cap && dis[edge[i].to]==-1)
                    dis[edge[i].to]=dis[now]+1,q.push(edge[i].to);
        }
    return dis[T]!=-1;
}
int dfs(int loc,int low)
{
    if (loc==T) return low;
    int w,used=0;
    for (int i=cur[loc]; i; i=edge[i].next)
        if (edge[i].cap && dis[edge[i].to]==dis[loc]+1)
            {
                w=dfs(edge[i].to,min(low-used,edge[i].cap));
                edge[i].cap-=w; edge[i^1].cap+=w; used+=w;
                if (edge[i].cap) cur[loc]=i; if (used==low) return low;
            }
    if (!used) dis[loc]=-1;
    return used;
}
#define inf 0x7fffffff
int dinic()
{
    int tmp=0;
    while (bfs())
        {
            for (int i=S; i<=T; i++) cur[i]=head[i];
            tmp+=dfs(S,inf);
        }
    return tmp;
}
void Build()
{
    S=0,T=N+1;
    for (int i=1; i<=N; i++) 
        if ((A[i]%2)) insert(S,i,B[i]); else insert(i,T,B[i]);
    for (int i=1; i<=N; i++)
        for (int j=1; j<=N; j++)    
            if ((A[i]%2) && !(A[j]%2) && !check(A[i],A[j]))
                insert(i,j,inf);
}
int main()
{
    N=read();
    for (int i=1; i<=N; i++) A[i]=read();
    for (int i=1; i<=N; i++) B[i]=read(),tot+=B[i];
    Build();
    int maxflow=dinic();
    printf("%d\n",tot-maxflow);
    return 0;
}

BZOJ-3158

【BZOJ-3275&3158】Number&千钧一发 最小割

3275: Number

Description

Input

Output

Sample Input

Sample Output

HINT

Source

3158: 千钧一发

Description

Input

Output

Sample Input

Sample Output

HINT

Source

Solution

Code

【BZOJ-3275&3158】Number&千钧一发最小割