POJ3068 "Shortest" pair of paths

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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;

}

巴特西

POJ3068 "Shortest" pair of paths

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