POJ 3204 Ikki's Story I - Road Reconstruction 残余网络

题意：给你一个有向图 问你跑完最大流之后可以通过改变多少条边使得最大流流量增加

思路：能够改变流量的边一定是最小割上的边 因此要找到最小割 我们的做法是找到满流的边 但是如果一条增广路上不止一条边满流 那么改变这条边的容量是没有意义的 因为另一条满流边还是会限制流量的增加 正确做法是 不仅要从源点找满流边 还要从汇点逆向找 正向访问过的点标记为1 逆向访问过的点标记为2 如果一条边的起点是1 终点是2那么这条边才满足题意

#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 = 500 + 10;
const int maxe = 10000 + 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;
          int vis[maxn], ans;
          Edge edge[maxe];
          void Init(int n){
                    this -> n = n;
                    cntE = ans = 0;
                    CLR(Head, -1);
                    CLR(vis, 0);
          }
          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]] ++;
                              }
                    }
          }
          void 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;
          }
          void Findcut1(int x){
                    vis[x] = 1;
                    for(int i = Head[x]; ~i; i = edge[i].next){
                              if(edge[i].c > edge[i].f && !vis[edge[i].v])
                                        Findcut1(edge[i].v);
                    }
          }
          void Findcut2(int x){
                    vis[x] = 2;
                    for(int i = Head[x]; ~i; i = edge[i].next){
                              if(edge[i^1].c > edge[i^1].f && !vis[edge[i].v])
                                        Findcut2(edge[i].v);
                    }
          }
          int Ans(){
                    REP(u, 0, n) for(int i = Head[u]; ~i; i = edge[i].next){
                              Edge &e = edge[i];
                              if(e.c && vis[u] == 1 && vis[e.v] == 2) ans++;
                    }
                    return ans;
          }
}solver;

void solve(){
          int n, m;
          scanf("%d%d", &n, &m);
          solver.Init(n);
          REP(i, 0, m){
                    int u, v, d;
                    scanf("%d%d%d", &u, &v, &d);
                    solver.Add(u, v, d);
          }
          solver.Maxflow(0, n - 1);
          solver.Findcut1(0); solver.Findcut2(n - 1);
          printf("%d\n", solver.Ans());
}

int main()
{
    //freopen("in.txt", "r", stdin);
    solve();
    return 0;
}