• 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
    ", l);
      }
      return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/dwtfukgv/p/7755727.html
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