POJ 3621 Sightseeing Cows (bellman-Ford + 01分数规划)

题意：给出 n 个点 m 条有向边，要求选出一个环，使得这上面点权和/边权和最大。

析：同样转成是01分数规划的形式，F / L 要这个值最大，也就是 G(r) = F - L * r 这个值为0时，r 的值，然后对于 F > 0，很明显是 r 太小，但是不好判断，把这个值取反，这样的话就能用Bellan-Ford 来判是不是有负环了，也可以用spfa。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <cstdio>

#include <string>

#include <cstdlib>

#include <cmath>

#include <iostream>

#include <cstring>

#include <set>

#include <queue>

#include <algorithm>

#include <vector>

#include <map>

#include <cctype>

#include <cmath>

#include <stack>

#include <sstream>

#include <list>

#include <assert.h>

#include <bitset>

#include <numeric>

#define debug() puts("++++")

#define gcd(a, b) __gcd(a, b)

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define fi first

#define se second

#define pb push_back

#define sqr(x) ((x)*(x))

#define ms(a,b) memset(a, b, sizeof a)

#define sz size()

#define pu push_up

#define pd push_down

#define cl clear()

#define all 1,n,1

#define FOR(i,x,n)  for(int i = (x); i < (n); ++i)

#define freopenr freopen("in.txt", "r", stdin)

#define freopenw freopen("out.txt", "w", stdout)

using namespace std;

typedef long long LL;

typedef unsigned long long ULL;

typedef pair<LL, int> P;

const int INF = 0x3f3f3f3f;

const LL LNF = 1e17;

const double inf = 1e20;

const double PI = acos(-1.0);

const double eps = 1e-3;

const int maxn = 1e3 + 10;

const int maxm = 1e6 + 5;

const int mod = 1000000007;

const int dr[] = {-1, 0, 1, 0};

const int dc[] = {0, -1, 0, 1};

const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};

int n, m;

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

inline bool is_in(int r, int c) {

  return r >= 0 && r < n && c >= 0 && c < m;

}

struct Edge{

  int to, val, next;

};

Edge edges[maxn*5];

int head[maxn], cnt;

void addEdge(int u, int v, int c){

  edges[cnt].to = v;

  edges[cnt].val = c;

  edges[cnt].next = head[u];

  head[u] = cnt++;

}

int val[maxn];

bool inq[maxn];

int num[maxn];

double d[maxn];

bool judge(double m){

  queue<int> q;  q.push(1);

  ms(inq, 0);  ms(num, 0);

  for(int i = 0; i <= n; ++i)  d[i] = inf;

  d[1] = 0;  inq[1] = true;

  while(!q.empty()){

    int u = q.front();  q.pop();

    inq[u] = 0;

    for(int i = head[u]; ~i; i = edges[i].next){

      int v = edges[i].to;

      if(d[v] > -val[v] + edges[i].val * m + d[u]){

        d[v] = -val[v] + edges[i].val * m + d[u];

        if(!inq[v]){ inq[v] = 1;  q.push(v);  if(++num[v] > n)  return true; }

      }

    }

  }

  return false;

}

int main(){

  while(scanf("%d %d", &n, &m) == 2){

    for(int i = 1; i <= n; ++i)  scanf("%d", val + i);

    ms(head, -1);  cnt = 0;

    while(m--){

      int u, v, c;

      scanf("%d %d %d", &u, &v, &c);

      addEdge(u, v, c);

    }

    double l = 0.0, r = 1e3;

    while(r - l > eps){

      double m = (l + r) / 2.0;

      judge(m) ? l = m : r = m;

    }

    printf("%.2f\n", l);

  }

  return 0;

}

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POJ 3621 Sightseeing Cows (bellman-Ford + 01分数规划)

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