LA 6395 SurelyYouCongest 最大流

题意：一个城市有n个路口m条路，每一条路一次性只能让一个人通过，否则就会形成阻塞。但是在路口的人数是不限的，现在给出一些人的位置，他们同时出发赶往路口1，同时他们只走最短的路，问你在不形成阻塞的前提下最多有多少人能到达路口1。

思路：
1：求出1到其他路口的最短路，然后只保留最短路。
2：只有当人在到路口1的距离相同的路口才有可能形成阻塞。
3：对人到路口距离排序，对于相同距离的人跑最大流，一层一层累加即可。

这题刚开始的思路很好出，但是往往会忽略思路2，导致WA。还有对于每一层如果不是现建图，现跑最大流，而是先建好图再一层一层累加的话会超时。不愧是Final题~

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
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....")

const int maxn = 25000 + 10;
const int maxe = 110000 + 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 visit[maxn];
  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;

struct E{
    int u, d;
    int next;
    E(int u = 0, int d = 0, int next = 0) : u(u), d(d), next(next) {}
    bool operator < (const E &rhs) const{
        return d > rhs.d;
    }
};

const int inf = 0x3f3f3f3f;
int H[maxn], cnte;
int vis[maxn], dist[maxn];
Edge edges[maxe];

void dijkstra(int s){
    priority_queue<E> Q;
    Q.push(E(s, 0));
    memset(vis, 0, sizeof(vis));
    memset(dist, inf, sizeof(dist));
    dist[s] = 0;
    while(!Q.empty()){
        E cur = Q.top(); Q.pop();
        int u = cur.u;
        int d = cur.d;
        if(vis[u]) continue;
        vis[u] = 1;
        for(int i = H[u]; ~i; i = edges[i].next){
            int v = edges[i].v;
            int d = edges[i].c;
            if(dist[v] > dist[u] + d){
                dist[v] = dist[u] + d;
                Q.push(E(v, dist[v]));
            }
        }
    }
}

void init(){
    memset(H, -1, sizeof(H));
    cnte = 0;
}

void Add(int u, int v, int d){
    edges[cnte] = Edge(v, d, 0, H[u]);
    H[u] = cnte++;
    edges[cnte] = Edge(u, d, 0, H[v]);
    H[v] = cnte++;
}

void build_graph(int u){
    if(vis[u]) return;
    vis[u] = 1;
    for(int i = H[u]; ~i; i = edges[i].next){
        int v = edges[i].v;
        int d = edges[i].c;
        if(dist[v] + d == dist[u]){
            solver.Add(u, v, 1);
            build_graph(v);
        }
    }
}

bool cmp(int a, int b){
    return dist[a] < dist[b];
}

int n, m, c;
int pos[1010];

int find_same(int k){
    int len = 0;
    for(int i = k; i < c; i++){
        if(dist[pos[i]] == dist[pos[k]]) len++;
        else break;
    }
    return k + len - 1;
}

void solve(){
    int S = 0, T = 1;
    init();
    for(int i = 0; i < m; i++){
        int u, v, d;
        scanf("%d%d%d", &u, &v, &d);
        Add(u, v, d);
    }
    for(int i = 0; i < c; i++)
        scanf("%d", &pos[i]);
    dijkstra(1);
    int ans = 0;
    sort(pos, pos + c, cmp);
    for(int i = 0, j = 0; i < c; i = j + 1){
        j = find_same(i);
        solver.Init(n + 1);
        memset(vis,0, sizeof(vis));
        for(int k = i; k <= j; k++){
            solver.Add(S, pos[k], 1);
            build_graph(pos[k]);
        }
        ans += solver.Maxflow(S, T);
    }
    printf("%d\n", ans);
}

int main()
{
    //freopen("in.txt", "r", stdin);
    while(~scanf("%d%d%d", &n, &m ,&c)) solve();
    return 0;
}