BZOJ2661 [BeiJing wc2012]连连看

把每个数拆成两个点建图

具体原因我想了想。。。因为一个点一定是不能做的。。。但是两个点不能保证一定是对称流法啊。。。（坑）

如果两个数a, b满足要求，则a -> b', b -> a'，边流量为1，费用为- a - b

最后再建源汇S, T，分别连边，流量为1，费用为0

跑一边费用流即可，但是要记下流量

/**************************************************************
    Problem: 2661
    User: rausen
    Language: C++
    Result: Accepted
    Time:128 ms
    Memory:3984 kb
****************************************************************/
 
#include <cstdio>
#include <cmath>
#include <algorithm>
 
using namespace std;
const int N = 2005;
const int M = N * 100;
const int inf = (int) 1e9;
 
struct edges {
    int next, to, f, cost;
    edges() {}
    edges(int _n, int _t, int _f, int _c) : next(_n), to(_t), f(_f), cost(_c) {}
} e[M];
  
int n, S, T;
int first[N], tot = 1;
int d[N], g[N], q[N];
bool v[N];
 
   
inline void Add_Edges(int x, int y, int f, int c) {
    e[++tot] = edges(first[x], y, f, c), first[x] = tot;
    e[++tot] = edges(first[y], x, 0, -c), first[y] = tot;
}
  
inline int calc() {
    int flow = inf, x;
    for (x = g[T]; x; x = g[e[x ^ 1].to])
        flow = min(flow, e[x].f);
    for (x = g[T]; x; x = g[e[x ^ 1].to])
        e[x].f -= flow, e[x ^ 1].f += flow;
    return flow;
}
  
bool spfa() {
    int x, y, l, r;
    for (x = 1; x <= T; ++x)
        d[x] = inf;
    d[S] = 0, v[S] = 1, q[0] = S;
    for(l = r = 0; l != (r + 1) % N; ++l %= N) {
        for (x = first[q[l]]; x; x = e[x].next)
            if (d[q[l]] + e[x].cost < d[y = e[x].to] && e[x].f) {
                d[y] = d[q[l]] + e[x].cost, g[y] = x;
                if (!v[y])
                    q[++r %= N] = y, v[y] = 1;
            }
        v[q[l]] = 0;
    }
    return d[T] != inf;
}
 
inline void work() {
    int ans1 = 0, ans2 = 0, del;
    while (spfa()) {
        del = calc();
        ans1 += del, ans2 += del * d[T];
    }
    printf("%d %d
", ans1 >> 1, -ans2 >> 1);
}
 
inline int sqr(int x) {
    return x * x;
}
 
inline int gcd(int a, int b) {
    return b ? gcd(b, a % b) : a;
}
 
inline bool check(int a, int b) {
    int t = sqr(b) - sqr(a);
    if (sqr((int) sqrt(t)) != t) return 0;
    return gcd(a, (int) sqrt(t)) == 1;
}
 
int main() {
    int a, b, i, j;
    scanf("%d%d", &a, &b);
    S = (b - a + 1) << 1 | 1, T = S + 1;
    for (i = a; i <= b; ++i)
        for (j = a; j < i; ++j) if (check(j, i))
            Add_Edges(i - a + 1, j + b - a * 2 + 2, 1, - i - j),
            Add_Edges(j - a + 1, i + b - a * 2 + 2, 1, - i - j);
    for (i = a; i <= b; ++i)
        Add_Edges(S, i - a + 1, 1, 0), Add_Edges(i + b - a * 2 + 2, T, 1, 0);
    work();
    return 0;
}