标签:efi memset 处理 com cto return cpp algo 最大
题意:给定一个 n 个点和相应的权值,要求你用 n-1 条边连接起来,其中一条边是魔法边,不用任何费用,其他的边是长度,求该魔法边的两端的权值与其他边费用的尽量大。
析:先求出最小生成树,然后再枚举每一条边,求出最大值,任意两点之间的距离可以通过预处理来解决,最小生成树时,要用prime算法,要不然可能会超时。
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> #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 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<int, int> P; const int INF = 0x3f3f3f3f; const double inf = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 1e3 + 10; const LL mod = 1e9 + 7; 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; } int cost[maxn][maxn]; int lowc[maxn]; int x[maxn], y[maxn], z[maxn]; bool vis[maxn]; int pre[maxn]; int dp[maxn][maxn]; vector<int> G[maxn]; void add(int u, int v){ G[u].push_back(v); G[v].push_back(u); } int dist(int i, int j){ return ((x[i] - x[j]) * (x[i] - x[j]) + (y[i] - y[j]) * (y[i] - y[j])); } double solve(){ double res = 0; memset(vis, 0, sizeof vis); memset(pre, -1, sizeof pre); vis[0] = 1; int p; int minc; int pr = 0; for(int i = 1; i < n; ++i) lowc[i] = cost[0][i]; for(int i = 1; i < n; ++i){ minc = INF; p = -1; for(int j = 0; j < n; ++j) if(!vis[j] && minc > lowc[j]){ minc = lowc[j]; p = j; } for(int j = 0; j < n; ++j) if(vis[j]){ if(minc == cost[p][j]){ pre[p] = j; break; } } res += sqrt(minc); vis[p] = 1; for(int j = 0; j < n; ++j) if(!vis[j] && lowc[j] > cost[p][j]) lowc[j] = cost[p][j]; } return res; } void dfs(int u, int fa, int rt, int mmax){ for(int i = 0; i < G[u].size(); ++i){ int v = G[u][i]; if(v == fa) continue; dp[rt][v] = max(mmax, cost[u][v]); dfs(v, u, rt, max(mmax, cost[u][v])); } } int main(){ int T; cin >> T; while(T--){ scanf("%d", &n); for(int i = 0; i < n; ++i){ scanf("%d %d %d", x+i, y+i, z+i); G[i].clear(); } for(int i = 0; i < n; ++i) for(int j = i+1; j < n; ++j) cost[i][j] = cost[j][i] = dist(i, j); double sum = solve(); for(int i = 0; i < n; ++i){ if(pre[i] == -1) continue; add(i, pre[i]); } memset(dp, 0, sizeof dp); for(int i = 0; i < n; ++i) dfs(i, -1, i, 0); double ans = 0.0; for(int i = 0; i < n; ++i) for(int j = i+1; j < n; ++j){ double x = z[i] + z[j]; double t = sum - sqrt(dp[i][j]); double xxx = x / t; ans = max(ans, xxx); } printf("%.2f\n", ans); } return 0; }
HDU 4081 Peach Blossom Spring (最小生成树+dfs)
标签:efi memset 处理 com cto return cpp algo 最大
原文地址:http://www.cnblogs.com/dwtfukgv/p/7241521.html