POJ 3469 Dual Core CPU 最小割

来源：互联网发布：中国移动宽带拨号软件编辑：程序博客网时间：2024/05/16 19:41

题意：给你一个双核处理器再给你n个进程在这两个处理器上的处理时间但是这些进程需要相互交换信息给你m行信息u v c 分别代表如果u v进程不在同一个处理器交换信息则要花费额外的时间c 问你运行完这n个进程的最小时间花费

思路：对于一个事物有两个状态的一般要用到二分图或者最小割来解决这题就是最小割建图方法是对于进程i在两个处理器的花费时间a b 分别建图S到i 容量为b 代表如果想让i属于T集合则要花费b来割断这条边然后建图i到T 容量为a 意义相似然后对于限制条件u v c 分别建图u到v v到u 容量均为c 代表如果想要u v 不属于容易个集合则要花费c来割断这条边然后跑最大流即可

#include <cstdio>#include <cstring>#include <algorithm>#include <vector>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 = 20000 + 10;const int maxe = 1000000 + 10;const int inf = 1e9;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 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]] ++;                              }                    }          }          int Maxflow(int s, int t){                    this -> s = s;                    this -> t = t;                    CPY(cur, Head);                    Bfs();                    int u = p[s] = s, 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 += 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, m;void solve(){          int S = 0, T = n + 1;          solver.Init(T + 1);          FOR(i, 1, n){                    int a, b;                    scanf("%d%d", &a, &b);                    solver.Add(S, i, b);                    solver.Add(i, T, a);          }          REP(i, 0, m){                    int u, v, d;                    scanf("%d%d%d", &u, &v, &d);                    solver.Add(u, v, d);                    solver.Add(v, u, d);          }          printf("%d\n", solver.Maxflow(S, T));}int main(){          //freopen("in.txt", "r", stdin);          while(~scanf("%d%d", &n, &m)) solve();          return 0;}

0 0