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今年的最后一篇了呢。。。好伤感的说,2014年还有1h就过去了
不不不回到正题,这道题嘛~看上去好神啊!
看到此题,我们可以联想到最优比例MST,于是就有了方法:
首先二分答案ans,判断ans是否可行,那如何判断呢?
每条边边权 - ans,之后在新的图中找负环即可。(可以用dfs版的spfa)
1 /************************************************************** 2 Problem: 1486 3 User: rausen 4 Language: C++ 5 Result: Accepted 6 Time:284 ms 7 Memory:1156 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 #include <cstring> 12 13 using namespace std; 14 typedef double lf; 15 const int N = 3005; 16 const int M = 10005; 17 const int inf = 1e9; 18 const lf eps = 1e-10; 19 20 inline int read() { 21 int x = 0, sgn = 1; 22 char ch = getchar(); 23 while (ch < ‘0‘ || ‘9‘ < ch) { 24 if (ch == ‘-‘) sgn = -1; 25 ch = getchar(); 26 } 27 while (‘0‘ <= ch && ch <= ‘9‘) { 28 x = x * 10 + ch - ‘0‘; 29 ch = getchar(); 30 } 31 return sgn * x; 32 } 33 34 struct Edge { 35 int x, y; 36 lf v; 37 38 inline void Read() { 39 x = read(), y = read(); 40 scanf("%lf", &v); 41 } 42 } E[M]; 43 44 struct edge { 45 int next, to; 46 lf v; 47 edge() {} 48 edge(int _n, int _t, lf _v) : next(_n), to(_t), v(_v) {} 49 } e[M]; 50 51 int n, m; 52 int first[N], tot; 53 lf d[N]; 54 bool f, v[N]; 55 56 inline void add_edge(int x, int y, lf v) { 57 e[++tot] = edge(first[x], y, v); 58 first[x] = tot; 59 } 60 61 void spfa(int p) { 62 int x, y; 63 v[p] = 1; 64 for (x = first[p]; x; x = e[x].next) 65 if (d[p] + e[x].v < d[y = e[x].to]) { 66 if (v[y]) { 67 f = 1; 68 break; 69 } else 70 d[y] = d[p] + e[x].v, spfa(y); 71 } 72 v[p] = 0; 73 } 74 75 void build_graph(lf x) { 76 int i; 77 tot = 0, memset(first, 0, sizeof(first)); 78 for (i = 1; i <= m; ++i) 79 add_edge(E[i].x, E[i].y, E[i].v - x); 80 } 81 82 bool check(lf x) { 83 int i; 84 build_graph(x); 85 memset(v, 0, sizeof(v)), memset(d, 0, sizeof(d)); 86 for (i = 1, f = 0; i <= n; ++i) { 87 spfa(i); 88 if (f) return 1; 89 } 90 return 0; 91 } 92 93 int main() { 94 int i; 95 scanf("%d%d", &n, &m); 96 for (i = 1; i <= m; ++i) 97 E[i].Read(); 98 lf l = -inf, r = inf, mid; 99 while (l + eps < r) { 100 mid = (l + r) / 2; 101 if (check(mid)) r = mid; else l = mid; 102 } 103 printf("%.8lf\n", l); 104 return 0; 105 }
(p.s. 这就是神奇的叫"分数规划"的东西!!!)
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原文地址:http://www.cnblogs.com/rausen/p/4196757.html