ZOJ 3760 Treasure Hunting 二分图带权最大独立集 最小割

题意：给你一个偶数P和二维坐标上的一些点(x, y) 权值为 x&y 让你选出其中一些点 使得权值和最大 且选中的点集中不存在这样的两点：(a, b)使得gcd(a.x^a.y^b.x^b.y, p) <= 1 

思路：由于题意中已经明确说明P为偶数 而两个偶数的最大公约数永远大于1 又发现对于任意两个数A B 只要它俩奇偶性相同 则 A^B必为偶数！所以我们只需要把图中所有点的权值按照奇偶性划分 它的原型不就是二分图带权最大独立集么！！建图方法

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;#define REP( i, a, b ) for( int i = a; i < b; i++ )#define FOR( i, a, b ) for( int i = a; i <= b; i++ )#define CLR( a, x ) memset( a, x, sizeof a )#define CPY( a, x ) memcpy( a, x, sizeof a )#define BUG puts( "**** BUG ****" )typedef long long LL;const int maxn = 500 + 10;const int maxe = 250000 + 10;const int INF = 0x7fffffff;struct Edge{          int v, c, f;          int next;          Edge() {}          Edge(int v, int c, int f, int next) : v(v), c(c), f(f), next(next) {}};struct Point{          int x, y;          Point(int x = 0, int y = 0) : x(x), y(y) {}}point[maxn];struct ISAP{          int n, s, t;          int num[maxn], cur[maxn], d[maxn], p[maxn];          int Head[maxn], cntE;          int Q[maxn], head, tail;          Edge edge[maxe];          void Init(int n){                    this -> n = n;                    cntE = 0;                    CLR(Head, -1);          }          void Add(int u, int v, int c){                    edge[cntE] = Edge(v, c, 0, Head[u]);                    Head[u] = cntE++;                    edge[cntE] = Edge(u, 0, 0, Head[v]);                    Head[v] = cntE++;          }          void Bfs(){                    CLR(d, -1);                    CLR(num, 0);                    d[t] = 0;                    head = tail = 0;                    Q[tail++] = t;                    num[0] = 1;                    while(head != tail){                              int u = Q[head++];                              for(int i = Head[u]; ~i; i = edge[i].next){                                        Edge &e = edge[i];                                        if(~d[e.v]) continue;                                        d[e.v] = d[u] + 1;                                        Q[tail++] = e.v;                                        num[d[e.v]] ++;                              }                    }          }          LL Maxflow(int s, int t){                    this -> s = s;                    this -> t = t;                    CPY(cur, Head);                    Bfs();                    int u = p[s] = s;                    LL flow = 0;                    while(d[s] < n){                              if(u == t){                                        int f = INF, neck;                                        for(int i = s; i != t; i = edge[cur[i]].v){                                                  if(f > edge[cur[i]].c - edge[cur[i]].f){                                                            f = edge[cur[i]].c - edge[cur[i]].f;                                                            neck = i;                                                  }                                        }                                        for(int i = s; i != t; i = edge[cur[i]].v){                                                  edge[cur[i]].f += f;                                                  edge[cur[i]^1].f -= f;                                        }                                        flow += (LL)f;                                        u = neck;                              }                              int ok = 0;                              for(int i = cur[u]; ~i; i = edge[i].next){                                        Edge &e = edge[i];                                        if(e.c > e.f && d[e.v] + 1 == d[u]){                                                  ok = 1;                                                  cur[u] = i;                                                  p[e.v] = u;                                                  u = e.v;                                                  break;                                        }                              }                              if(!ok){                                        int m = n - 1;                                        if(--num[d[u]] == 0) break;                                        for(int i = Head[u]; ~i; i = edge[i].next){                                                  Edge &e = edge[i];                                                  if(e.c - e.f > 0 && m > d[e.v]){                                                            cur[u] = i;                                                            m = d[e.v];                                                  }                                        }                                        ++num[d[u] = m + 1];                                        u = p[u];                              }                    }                    return flow;          }}solver;int n, p;Point read_Point(){          Point a;          scanf("%d%d", &a.x, &a.y);          return a;}int gcd(int a, int b){ return b == 0 ? a : gcd(b, a % b); }void input(){ FOR(i, 1, n) point[i] = read_Point(); }void solve(){          int S = 0, T = n + 1;          LL sum = 0;          solver.Init(n+2);          FOR(i, 1, n){                    if((point[i].x ^ point[i].y) & 1){                              solver.Add(S, i, point[i].x & point[i].y);                              FOR(j, 1, n){                              if(!((point[j].x^point[j].y)&1)&&gcd(point[i].x^point[i].y^point[j].x^point[j].y, p)<=1)                                        solver.Add(i, j, INF);                              }                    }                    else solver.Add(i, T, point[i].x & point[i].y);                    sum += (LL)(point[i].x & point[i].y);          }          printf("%lld\n", sum - solver.Maxflow(S, T));}int main(){          //freopen("in.txt", "r", stdin);          while(~scanf("%d%d", &n, &p)){                    input();                    solve();          }          return 0;}