[AHOI2008]上学路线

嘟嘟嘟

这道题其实挺显然的。

首先dijkstra跑出最短路图，然后在最短路图上求最小割。

正确性显然。

需要注意的是，在新图中添加最短路图的边的时候，一定是跑完dijkstra再加边，如果边跑dijkstra边加边，得到的是最短路树，而不是图。我因为这个WA了好几发。

#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 rg register
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 505;
const int maxe = 1.3e5 + 5;
inline ll read()
{
  ll ans = 0;
  char ch = getchar(), last = ' ';
  while(!isdigit(ch)) {last = ch; ch = getchar();}
  while(isdigit(ch)) {ans = ans * 10 + 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;
struct Edge1
{
  int nxt, to, t, c;
}e1[maxe << 1];
int head1[maxn], ecnt1 = -1;
void addEdge1(int x, int y, int t, int c)
{
  e1[++ecnt1] = (Edge1){head1[x], y, t, c};
  head1[x] = ecnt1;
}

struct Edge2
{
  int nxt, from, to, cap, flow;
}e2[maxe << 2];
int head2[maxn], ecnt2 = -1;
void addEdge2(int from, int to, int w)
{
  e2[++ecnt2] = (Edge2){head2[from], from, to, w, 0};
  head2[from] = ecnt2;
  e2[++ecnt2] = (Edge2){head2[to], to, from, 0, 0};
  head2[to] = ecnt2;
}

#define pr pair<int, int>
#define mp make_pair
bool in[maxn];
int dis[maxn];
void dijkstra(int s)
{
  Mem(dis, 0x3f); dis[s] = 0;
  priority_queue<pr, vector<pr>, greater<pr> > q;
  q.push(mp(dis[s], s));
  while(!q.empty())
    {
      int now = q.top().second; q.pop();
      if(in[now]) continue;
      in[now] = 1;
      for(int i = head1[now]; i != -1; i = e1[i].nxt)
    {
      if(dis[e1[i].to] > dis[now] + e1[i].t)
        {
          dis[e1[i].to] = dis[now] + e1[i].t;
          q.push(mp(dis[e1[i].to], e1[i].to));
        }
    }
    }
}

int dis2[maxn];
bool bfs()
{
  Mem(dis2, 0); dis2[1] = 1;
  queue<int> q; q.push(1);
  while(!q.empty())
    {
      int now = q.front(); q.pop();
      for(int i = head2[now]; i != -1; i = e2[i].nxt)
    {
      if(!dis2[e2[i].to] && e2[i].cap > e2[i].flow)
        {
          dis2[e2[i].to] = dis2[now] + 1;
          q.push(e2[i].to);
        }
    }
    }
  return dis2[n];
}
int cur[maxn];
int dfs(int now, int res)
{
  if(now == n || res == 0) return res;
  int flow = 0, f;
  for(int &i = cur[now]; i != -1; i = e2[i].nxt)
    {
      if(dis2[e2[i].to] == dis2[now] + 1 && (f = dfs(e2[i].to, min(res, e2[i].cap - e2[i].flow))) > 0)
    {
      e2[i].flow += f; e2[i ^ 1].flow -= f;
      flow += f; res -= f;
      if(res == 0) break;
    }
    }
  return flow;
}

int minCut()
{
  int flow = 0;
  while(bfs())
    {
      memcpy(cur, head2, sizeof(head2));
      flow += dfs(1, INF);
    }
  return flow;
}

int main()
{
  Mem(head1, -1); Mem(head2, -1);
  n = read(); m = read();
  for(int i = 1; i <= m; ++i)
    {
      int x = read(), y = read(), t = read(), c = read();
      addEdge1(x, y, t, c); addEdge1(y, x, t, c);
    }
  dijkstra(1);
  for(int i = 1; i <= n; ++i)
    for(int j = head1[i]; j != -1; j = e1[j].nxt)
      if(dis[e1[j].to] == dis[i] + e1[j].t) addEdge2(i, e1[j].to, e1[j].c);
  write(dis[n]), enter, write(minCut()), enter;
  return 0;
}