POJ3068 "Shortest" pair of paths

嘟嘟嘟




题目大意：一个有向图，每一条边有一个边权，求从节点\(0\)到\(n - 1\)的两条不经过同一条边的路径，并且边权和最小。




费用流板子题。

发个博客证明一下我写了这题。

#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 = 70;
const int maxm = 1e4 + 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, s, t;
struct Edge
{
  int nxt, from, to, cap, c;
}e[maxm << 1];
int head[maxn], ecnt = -1;
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;
}

queue<int> q;
bool in[maxn];
int dis[maxn], pre[maxn], flow[maxn];
bool spfa()
{
  Mem(in, 0); Mem(dis, 0x3f);
  in[s] = 1; dis[s] = 0; flow[s] = INF;
  q.push(s);
  while(!q.empty())
    {
      int now = q.front(); q.pop(); in[now] = 0;
      for(int i = head[now], v; i != -1; i = e[i].nxt)
	{
	  v = e[i].to;
	  if(e[i].cap && dis[now] + e[i].c < dis[v])
	    {
	      dis[v] = dis[now] + e[i].c;
	      pre[v] = i;
	      flow[v] = min(flow[now], e[i].cap);
	      if(!in[v]) in[v] = 1, q.push(v);
	    }
	}
    }
  return dis[t] != INF;
}
int maxFlow = 0, minCost = 0;
void update()
{
  int x = t;
  while(x != s)
    {
      int i = pre[x];
      e[i].cap -= flow[t];
      e[i ^ 1].cap += flow[t];
      x = e[i].from;
    }
  maxFlow += flow[t]; minCost += dis[t] * flow[t];
}
void MCMF()
{
  while(spfa()) update();
}

void init()
{
  Mem(head, -1); ecnt = -1;
  maxFlow = minCost = 0;
}

int main()
{
  int T = 0;
  while(scanf("%d%d", &n, &m) != EOF && n && m)
    {
      init();
      s = 0; t = n + 1;
      for(int i = 1; i <= m; ++i)
	{
	  int x = read() + 1, y = read() + 1, c = read();
	  addEdge(x, y, 1, c);
	}
      addEdge(s, 1, 2, 0); addEdge(n, t, 2, 0);
      MCMF();
      printf("Instance #%d: ", ++T);
      if(maxFlow < 2) puts("Not possible");
      else write(minCost), enter;
    }
  return 0;
}