[SHOI2010]最小生成树

嘟嘟嘟


这道题的切入点在于模型的转化。


令第\(Lab\)条边连接的两个点分别为\(x, y\)。根据kruskal算法，我们排完序加边的时候，在执行第\(Lab\)条边之前，都要保证\(x, y\)不连通。这就很像最小割了。
所以我们把边权小于\(w_{Lab}\)的都拿来建图，那边权是啥咧？题中说的除了一条边其余的都减1，那不就相当于这条边自己加1了嘛。那么为了让图不连通，只需让每一条边的容量为\(w_{lab} - w_i + 1\)即可。

#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 = 505;
const int maxm = 805;
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, id;
struct Node
{
  int x, y, w;
}q[maxm];
struct Edge
{
  int nxt, from, to, cap, flow;
}e[maxm << 2];
int head[maxn], ecnt = -1;
In void addEdge(int x, int y, int w)
{
  e[++ecnt] = (Edge){head[x], x, y, w, 0};
  head[x] = ecnt;
  e[++ecnt] = (Edge){head[y], 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();
      for(int i = head[now], v; i != -1; 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 != -1; i = e[i].nxt)
    {
      v = e[i].to;
      if(dis[v] == 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 minCut()
{
  int flow = 0;
  while(bfs())
    {
      memcpy(cur, head, sizeof(head));
      flow += dfs(0, INF);
    }
  return flow;
}

int main()
{
  Mem(head, -1);
  n = read(), m = read(), id = read(); t = n + 1;
  for(int i = 1; i <= m; ++i) q[i].x = read(), q[i].y = read(), q[i].w = read();
  for(int i = 1; i <= m; ++i)
    if(q[i].w <= q[id].w && i != id)
      {
	addEdge(q[i].x, q[i].y, q[id].w - q[i].w + 1);
	addEdge(q[i].y, q[i].x, q[id].w - q[i].w + 1);
      }
  addEdge(0, q[id].x, INF), addEdge(q[id].y, t, INF);
  write(minCut()), enter;
  return 0;
}