# 网络流/最小割

Orz了Jiry_2神犇，蒟蒻网络流建模什么的完全不会啊T_T

按奇偶性来分组实在太巧妙了……然后相关的点之间连边表示只能选其一，来求最小割……

  1 /**************************************************************
2     Problem: 3275
3     User: Tunix
4     Language: C++
5     Result: Accepted
6     Time:1252 ms
7     Memory:1708 kb
8 ****************************************************************/
9
10 //BZOJ 3275
11 #include<cmath>
12 #include<vector>
13 #include<cstdio>
14 #include<cstring>
15 #include<cstdlib>
16 #include<iostream>
17 #include<algorithm>
18 #define rep(i,n) for(int i=0;i<n;++i)
19 #define F(i,j,n) for(int i=j;i<=n;++i)
20 #define D(i,j,n) for(int i=j;i>=n;--i)
21 #define pb push_back
22 using namespace std;
23 inline int getint(){
24     int v=0,sign=1; char ch=getchar();
25     while(ch<'0'||ch>'9'){ if (ch=='-') sign=-1; ch=getchar();}
26     while(ch>='0'&&ch<='9'){ v=v*10+ch-'0'; ch=getchar();}
27     return v*sign;
28 }
29 const int N=3100,M=30010,INF=~0u>>2;
30 typedef long long LL;
31 /******************tamplate*********************/
32 int n,m,a[3010],tot,ans;
33 struct edge{
34     int to,v;
35 };
36 int gcd(int a,int b){if (!b) return a;else return gcd(b,a%b);}
37 bool judge(int a,int b){
38     int s=a*a+b*b;
39     int q=int(sqrt(s));
40     if (q*q!=s) return 0;
41     if (gcd(a,b)!=1) return 0;
42     return 1;
43 }
44 struct Net{
45     edge E[M];
47     bool vis[110][110];
48     void ins(int x,int y,int v){
49         E[++cnt]=(edge){y,v};
51     }
52     void add(int x,int y,int v){
53         ins(x,y,v); ins(y,x,0);
54     }
55     int s,t,cur[N],d[N],Q[N];
56     void init(){
57         n=getint();cnt=1;
58         tot=ans=0;
59         s=0; t=n+1;
60         F(i,1,n){
61             a[i]=getint();tot+=a[i];
64         }
65         F(i,1,n) if (a[i]&1)
66             F(j,1,n) if ((a[j]&1)==0)
68     }
69     bool mklevel(){
70         memset(d,-1,sizeof d);
71         d[s]=0;
72         int l=0,r=-1;
73         Q[++r]=s;
74         while(l<=r){
75             int x=Q[l++];
77                 if (d[E[i].to]==-1 && E[i].v){
78                     d[E[i].to]=d[x]+1;
79                     Q[++r]=E[i].to;
80                 }
81         }
82         return d[t]!=-1;
83     }
84     int dfs(int x,int a){
85         if (x==t) return a;
86         int flow=0;
87         for(int &i=cur[x];i && flow<a;i=next[i])
88             if (E[i].v && d[E[i].to]==d[x]+1){
89                 int f=dfs(E[i].to,min(a-flow,E[i].v));
90                 E[i].v-=f;
91                 E[i^1].v+=f;
92                 flow+=f;
93             }
94         if (!flow) d[x]=-1;
95         return flow;
96     }
97     void Dinic(){
98         while(mklevel()){
100             ans+=dfs(s,INF);
101         }
102     }
103 }G1;
104 int main(){
105 #ifndef ONLINE_JUDGE
106     freopen("3275.in","r",stdin);
107     freopen("3275.out","w",stdout);
108 #endif
109     G1.init(); G1.Dinic();
110     printf("%d\n",tot-ans);
111     return 0;
112 }
View Code

## 3275: Number

Time Limit: 10 Sec  Memory Limit: 128 MB
Submit: 495  Solved: 217
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## Description

1:存在正整数C，使a*a+b*b=c*c
2:gcd(a,b)=1

5
3 4 5 6 7

22

n<=3000。

## Source

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posted @ 2015-03-15 09:42  Tunix  阅读(335)  评论(0编辑  收藏  举报