LA5713 Qin Shi Huang's National Road System

Posted 2020-10-22 嘒彼小星

题目大意：秦始皇要在n个城市之间修筑一条道路使得任意两个城市均可连通。有个道士可以用法力帮忙修一条路。秦始皇希望其他的道路总长B最短且用法术连接的两个城市的人口之和A尽量大，因此下令寻找一个A / B的最大方案。（转自http://blog.csdn.net/murmured/article/details/18865721，侵删）

题解：考虑修哪一条路，此时A确定，断掉这条路后，形成两个连通块，不难发现两个连通块都应是MST才能让B最短。此时B为两个连通块的MST权值和
这个过程相当于选了一条路u->v后，在整张图的MST上，把MST上u->v路径上最大边删掉。

其实不用枚举边，枚举u和v即可

空间开小了，RE了无数发。。。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cstdlib>
 5 #include <algorithm>
 6 #include <queue>
 7 #include <vector>
 8 #include <map>
 9 #include <string> 
10 #include <cmath> 
11 #define min(a, b) ((a) < (b) ? (a) : (b))
12 #define max(a, b) ((a) > (b) ? (a) : (b))
13 #define abs(a) ((a) < 0 ? (-1 * (a)) : (a))
14 template<class T>
15 inline void swap(T &a, T &b)
16 {
17     T tmp = a;a = b;b = tmp;
18 }
19 inline void read(int &x)
20 {
21     x = 0;char ch = getchar(), c = ch;
22     while(ch < ‘0‘ || ch > ‘9‘) c = ch, ch = getchar();
23     while(ch <= ‘9‘ && ch >= ‘0‘) x = x * 10 + ch - ‘0‘, ch = getchar();
24     if(c == ‘-‘) x = -x;
25 }
26 
27 const int INF = 0x3f3f3f3f;
28 const int MAXN = 2000 + 10;
29 const int MAXM = 1000000 + 10;
30 
31 int x[MAXM], y[MAXM], node[MAXM], n, m, t, u[MAXM], v[MAXM], cnt[MAXM], fa[MAXM], vis[MAXN];
32 double ma[MAXN][MAXN], w[MAXM];
33 struct Edge
34 {
35     int u,v,nxt;
36     double w;
37     Edge(int _u, int _v, double _w, int _nxt){u = _u;v = _v;w = _w;nxt = _nxt;}
38     Edge(){} 
39 }edge[MAXM << 1];
40 int head[MAXN], cntt;
41 inline void insert(int a, int b, double c)
42 {
43     edge[++cntt] = Edge(a,b,c,head[a]);
44     head[a] = cntt;
45 }
46 int find(int x){return x == fa[x] ? x : fa[x] = find(fa[x]);}
47 int cmp(int x, int y){return w[x] < w[y];}
48 
49 int tiaoshi;
50 void dfs(int x, int pre)
51 {
52     vis[x] = 1;
53     for(int pos = head[x];pos;pos = edge[pos].nxt)
54     {
55         int v = edge[pos].v;
56         if(v == pre) continue;
57         for(int i = 1;i <= n;++ i)
58             if(i == v || !vis[i]) continue;
59             else ma[i][v] =ma[v][i] = max(ma[i][v], max(ma[v][i], max(ma[x][i], max(ma[i][x], edge[pos].w))));
60         dfs(v, x);
61     }
62 }
63 int main()
64 {
65     read(t);
66     for(;t;--t)
67     {
68         tiaoshi = 0;
69         read(n);memset(ma, 0, sizeof(ma)), memset(vis, 0, sizeof(vis)), memset(head, 0, sizeof(head)), cntt = 0, m = 0;
70         for(int i = 1;i <= n;++ i) read(x[i]), read(y[i]), read(node[i]);
71         for(int i = 1;i <= n;++ i) 
72             for(int j = i + 1;j <= n;++ j)
73                 ++ m, u[m] = i, v[m] = j, w[m] = sqrt((x[i] - x[j]) * (x[i] - x[j]) + (y[i] - y[j]) * (y[i] - y[j])), cnt[m] = m, fa[m] = m;
74         std::sort(cnt + 1, cnt + 1 + m, cmp);
75         int tmp = 0;
76         double ans = 0;
77         double ans2 = 0;
78         for(int i = 1;i <= m;++ i)
79         {
80             int f1 = find(u[cnt[i]]), f2 = find(v[cnt[i]]);
81             if(f1 == f2) continue;
82             fa[f1] = f2;
83             insert(u[cnt[i]], v[cnt[i]], w[cnt[i]]), insert(v[cnt[i]], u[cnt[i]], w[cnt[i]]);
84             ans += w[cnt[i]];
85             ++ tmp;if(tmp == n - 1) break;
86         }
87         dfs(1, -1);
88         for(int i = 1;i <= n;++ i)
89             for(int j = i + 1;j <= n;++j)
90             {
91                 if(ans == ma[i][j]) continue;
92                 ans2 = max(ans2, (double)(node[i] + node[j]) / (ans - ma[i][j]));
93             }
94         printf("%.2lf\n", ans2);
95     }
96     return 0;
97 }

LA5713