HDU 4940 无源汇带上下界可行流

来源：互联网发布：ubuntu火狐下载的软件编辑：程序博客网时间：2024/06/05 05:15

真的好久没刷题了浪费了一个月的时间应付期末考试现在几乎废了各种坑队友十分不在状态希望这个暑假尽快找回！

题意：给你一个连通图每条有向边有两个值一个是断开这条有向边的费用一个是断开后新建一条无向边的费用问你是否存在一个集合S 使得这个集合与其补集T 破坏所有S集合到T集合的边代价和是X 然后修复T到S的边为无向边代价和是Y 满足Y<x 满足输出unhappy 否则输出happy

思路：可行流需要自己体会（其实是我真的不知道怎么说）

#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 )const int maxn = 200 + 10;const int maxe = 20000 + 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 flow = 0, u = p[s] = s;                    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 cas = 0;void solve(){          int n, m, S, T, ans = 0;          int M[maxn];          memset(M, 0, sizeof(M));          scanf("%d%d", &n, &m);          S = 0, T = n + 1;          solver.Init(T + 1);          REP(i, 0, m){                    int u, v, b, c;                    scanf("%d%d%d%d", &u, &v, &b, &c);                    M[v] += b;                    M[u] -= b;                    solver.Add(u, v, c);          }          FOR(i, 1, n){                    if(M[i] >= 0) solver.Add(S, i, M[i]), ans += M[i];                    else solver.Add(i, T, -M[i]);          }          printf("Case #%d: ", ++cas);          if(solver.Maxflow(S, T) == ans) printf("happy\n");          else printf("unhappy\n");}int main(){    //freopen("in.txt", "r", stdin);    int T;    scanf("%d", &T);    while(T--) solve();    return 0;}

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