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  • [SDOI2010]星际竞速

    嘟嘟嘟


    带权最小边覆盖?
    最小边覆盖可以用二分图解决,那带权怎么办?
    一时zz想不出来,看了一眼标签发现跑费用流就行。
    把每一个点拆成(i)(i'),源点向(i)连容量为1,费用为0的边,(i')向汇点连((1, 0))的边。然后如果(x)(y)有边((x < y)),从(x)(y)((1, c))的边。
    对于每一个点的点权,我刚开始想每一个点都向(i')((1, a[i]))的边,但这样会TLE,实际上只用从源点向(i')((1, a[i]))的边,因为根据网络流性质,我们可以直接合并这些边。

    #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 = 2e3 + 5;
    const int maxe = 1e7 + 5;
    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, t, a[maxn];
    struct Edge
    {
      int nxt, from, to, cap, cos;
    }e[maxe];
    int head[maxn], ecnt = -1;
    In void addEdge(int x, int y, int w, int c)
    {
      e[++ecnt] = (Edge){head[x], x, y, w, c};
      head[x] = ecnt;
      e[++ecnt] = (Edge){head[y], y, x, 0, -c};
      head[y] = ecnt;
    }
    
    bool in[maxn];
    int dis[maxn], pre[maxn], flow[maxn];
    In bool spfa()
    {
      Mem(dis, 0x3f), Mem(in, 0);
      dis[0] = 0, flow[0] = INF;
      queue<int> q; q.push(0);
      while(!q.empty())
        {
          int now = q.front(); q.pop(); in[now] = 0;
          for(int i = head[now], v; ~i; i = e[i].nxt)
    	{
    	  if(dis[v = e[i].to] > dis[now] + e[i].cos && e[i].cap > 0)
    	    {
    	      dis[v] = dis[now] + e[i].cos;
    	      pre[v] = i;
    	      flow[v] = min(flow[now], e[i].cap);
    	      if(!in[v]) q.push(v), in[v] = 1;
    	    }
    	}
        }
      return dis[t] ^ INF;
    }
    int minCost = 0;
    In void update()
    {
      int x = t;
      while(x)
        {
          int i = pre[x];
          e[i].cap -= flow[t];
          e[i ^ 1].cap += flow[t];
          x = e[i].from;
        }
      minCost += flow[t] * dis[t];
    }
    
    In int MCMF()
    {
      while(spfa()) update();
      return minCost;
    }
    
    int main()
    {
      //freopen("ha.in", "r", stdin);
      Mem(head, -1);
      n = read(), m = read(); t = n + n + 1;
      for(int i = 1; i <= n; ++i) a[i] = read();
      for(int i = 1; i <= n; ++i)
        {
          addEdge(0, i, 1, 0), addEdge(i + n, t, 1, 0);
          addEdge(0, i + n, 1, a[i]);
        }
      for(int i = 1; i <= m; ++i)
        {
          int x = read(), y = read(), c = read();
          if(x > y) swap(x, y);
          addEdge(x, y + n, 1, c);
        }
      write(MCMF()), enter;
      return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/mrclr/p/10799326.html
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