网络流题有Dinic板子还正是爽啊 ≥v≤~2333
首先我们把一个数字拆成2个点，连边规则：
（1）S向i连权为a[i]的边，i + n向T连权为a[i]的边
（2）有关系的点互相连边，权为inf
则答案是sigma(a[i]) - 最小割值
1 /************************************************************** 2 Problem: 3275 3 User: rausen 4 Language: C++ 5 Result: Accepted 6 Time:1560 ms 7 Memory:6760 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 #include <cmath> 12 #include <cstring> 13 #include <algorithm> 14 15 using namespace std; 16 const int N = 6005; 17 const int M = 500001; 18 const int inf = (int) 1e9; 19 20 struct edges{ 21 int next, to, f; 22 edges() {} 23 edges(int _next, int _to, int _f) : next(_next), to(_to), f(_f) {} 24 } e[M]; 25 26 int n, m, ans, S, T; 27 int first[N], tot = 1; 28 int q[N], a[N >> 1], d[N]; 29 30 inline int read() { 31 int x = 0; 32 char ch = getchar(); 33 while (ch < '0' || '9' < ch) 34 ch = getchar(); 35 while ('0' <= ch && ch <= '9') { 36 x = x * 10 + ch - '0'; 37 ch = getchar(); 38 } 39 return x; 40 } 41 42 inline void add_edge(int x, int y, int z){ 43 e[++tot] = edges(first[x], y, z); 44 first[x] = tot; 45 } 46 47 inline void Add_Edges(int x, int y, int z){ 48 add_edge(x, y, z); 49 add_edge(y, x, 0); 50 } 51 52 bool bfs(){ 53 memset(d, 0, sizeof(d)); 54 q[1] = S, d[S] = 1; 55 int l = 0, r = 1, x, y; 56 while (l < r){ 57 ++l; 58 for (x = first[q[l]]; x; x = e[x].next){ 59 y = e[x].to; 60 if (!d[y] && e[x].f) 61 q[++r] = y, d[y] = d[q[l]] + 1; 62 } 63 } 64 return d[T]; 65 } 66 67 int dinic(int p, int limit){ 68 if (p == T || !limit) return limit; 69 int x, y, tmp, rest = limit; 70 for (x = first[p]; x; x = e[x].next){ 71 y = e[x].to; 72 if (d[y] == d[p] + 1 && e[x].f && rest){ 73 tmp = dinic(y, min(rest, e[x].f)); 74 rest -= tmp; 75 e[x].f -= tmp, e[x ^ 1].f += tmp; 76 if (!rest) return limit; 77 } 78 } 79 if (limit == rest) d[p] = 0; 80 return limit - rest; 81 } 82 83 int Dinic(){ 84 int res = 0, x; 85 while (bfs()) 86 res += dinic(S, inf); 87 return res; 88 } 89 90 int gcd(int a, int b) { 91 return b ? gcd(b, a % b) : a; 92 } 93 94 inline int Sqr(int x) { 95 return x * x; 96 } 97 98 inline bool check(int x, int y) { 99 int t = x * x + y * y; 100 return Sqr(((int) sqrt(t))) == t && gcd(x, y) == 1; 101 } 102 103 int main() { 104 int i, j; 105 n = read(); 106 S = n << 1 | 1, T = S + 1; 107 for (i = 1; i <= n; ++i) { 108 a[i] = read(); 109 Add_Edges(S, i, a[i]); 110 Add_Edges(i + n, T, a[i]); 111 ans += a[i]; 112 } 113 for (i = 1; i <= n; ++i) 114 for (j = i + 1; j <= n; ++j) 115 if (check(a[i], a[j])) 116 Add_Edges(i, j + n, inf), Add_Edges(j, i + n, inf); 117 printf("%d ", ans - Dinic() / 2); 118 return 0; 119 }