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  • [JSOI2008]Blue Mary的旅行

    嘟嘟嘟


    (n)那么小,就知道是网络流。然后二分,按时间拆点。


    刚开始我看成所有航班一天只能起飞一次,纠结了好一会儿。但实际上是每一个航班单独考虑,互不影响。
    建图很显然,拆完点后每一个点的第(i)天向和他相连的点的第(i + 1)天连边,同时自己的第(i)天也要向第(i + 1)天连边。


    刚开始数组开小了,自己造了几组大数据才对。不过说来也奇怪,自己的随机数据跑出来都3、4秒,交上去却秒A了。可能是因为自己的答案都是1或2吧,出题人为了防止固输骗分,就没有造这种极端数据。

    #include<cstdio>
    #include<iostream>
    #include<cmath>
    #include<algorithm>
    #include<cstring>
    #include<cstdlib>
    #include<cctype>
    #include<vector>
    #include<stack>
    #include<queue>
    using namespace std;
    #define enter puts("") 
    #define space putchar(' ')
    #define Mem(a, x) memset(a, x, sizeof(a))
    #define In inline
    typedef long long ll;
    typedef double db;
    const int INF = 0x3f3f3f3f;
    const db eps = 1e-8;
    const int maxn = 62505;
    const int maxe = 1e6 + 5;
    const int maxm = 2455;
    inline ll read()
    {
      ll ans = 0;
      char ch = getchar(), last = ' ';
      while(!isdigit(ch)) last = ch, ch = getchar();
      while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
      if(last == '-') ans = -ans;
      return ans;
    }
    inline void write(ll x)
    {
      if(x < 0) x = -x, putchar('-');
      if(x >= 10) write(x / 10);
      putchar(x % 10 + '0');
    }
    
    int n, m, tot, t;
    struct Node
    {
      int x, y, w;
    }ed[maxm];
    
    struct Edge
    {
      int nxt, to, cap, flow;
    }e[maxe];
    int head[maxn], ecnt = -1;
    In void addEdge(int x, int y, int w)
    {
      //printf("()%d %d %d
    ", x, y, w);
      e[++ecnt] = (Edge){head[x], y, w, 0};
      head[x] = ecnt;
      e[++ecnt] = (Edge){head[y], x, 0, 0};
      head[y] = ecnt;
    }
    
    int dis[maxn];
    In bool bfs()
    {
      Mem(dis, 0); dis[0] = 1;
      queue<int> q; q.push(0);
      while(!q.empty())
        {
          int now = q.front(); q.pop();
          //printf("#%d
    ", now);
          for(int i = head[now], v; ~i; i = e[i].nxt)
    	{
    	  if(!dis[v = e[i].to] && e[i].cap > e[i].flow)
    	    {
    	      dis[v] = dis[now] + 1;
    	      q.push(v);
    	    }
    	}
        }
      return dis[t];
    }
    int cur[maxn];
    In int dfs(int now, int res)
    {
      if(now == t || res == 0) return res;
      int flow = 0, f;
      for(int& i = cur[now], v; ~i; i = e[i].nxt)
        {
          if(dis[v = e[i].to] == dis[now] + 1 && (f = dfs(v, min(res, e[i].cap - e[i].flow))) > 0)
    	{
    	  e[i].flow += f, e[i ^ 1].flow -= f;
    	  flow += f, res -= f;
    	  if(res == 0) break;
    	}
        }
      return flow;
    }
    In int maxflow()
    {
      int flow = 0;
      while(bfs())
        {
          memcpy(cur, head, sizeof(head));
          flow += dfs(0, INF);
        }
      return flow;
    }
    
    int _Max = 0;
    In int num(int x, int tim, int id) {return (x - 1) * tim + id;}
    In bool judge(int tim)
    {
      ecnt = -1; Mem(head, -1);
      t = num(n, tim, tim) + 1;
      _Max = max(_Max, t);
      for(int i = 1; i <= m; ++i)
        for(int j = 1; j < tim; ++j) addEdge(num(ed[i].x, tim, j), num(ed[i].y, tim, j + 1), ed[i].w);
      for(int i = 1; i <= tim; ++i)
        {
          addEdge(0, num(1, tim, i), INF);
          addEdge(num(n, tim, i), t, INF);
        }
      for(int i = 1; i <= n; ++i)
        for(int j = 2; j <= tim; ++j) addEdge(num(i, tim, j - 1), num(i, tim, j), INF);
      return maxflow() >= tot;
    }
    
    int main()
    {
      //freopen("ha.in", "r", stdin);
      Mem(head, -1);
      n = read(), m = read(), tot = read();
      for(int i = 1; i <= m; ++i) ed[i].x = read(), ed[i].y = read(), ed[i].w = read();
      int L = 0, R = tot * n + 1;
      while(L < R)
        {
          int mid = (L + R) >> 1;
          if(judge(mid)) R = mid;
          else L = mid + 1;
        }
      write(L - 1), enter;
      //write(_Max), enter;
      return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/mrclr/p/10788538.html
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