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; }