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;}