题目:https://cn.vjudge.net/problem/HDU-1875
题意:平面坐标系上,C个点,坐标[0, 1000]整数,要求修路使之变成连通图,一条边费用等于长度*100,长度不在[10, 1000]范围内的边不能选择。求最小费用。
思路:二重循环计算两点距离建边,边长不符合[10, 1000]的边忽略,之后求最小生成树即可。
代码:C++
#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <cstdlib>
#include <algorithm>
#include <string>
#include <stack>
#include <cmath>
using namespace std;
const int maxn = 100 + 10;
struct Edge
{
int from, to;
double val;
Edge(int from, int to, double val): from(from), to(to), val(val) {}
bool operator < (const Edge &t) const
{
return val < t.val;
}
};
typedef pair<int, int> pr;
int n;
vector<pr> vp;
vector<Edge> ve;
int fa[maxn];
int findset(int x)
{
return fa[x] == x ? x : fa[x] = findset(fa[x]);
}
void unionset(int x, int y)
{
int p1 = findset(x);
int p2 = findset(y);
if(p1 != p2)
{
fa[p1] = p2;
}
}
int main()
{
int T;
cin >> T;
while(T--)
{
ve.clear();
vp.clear();
scanf("%d", &n);
for(int i = 0; i < n; i++)
{
fa[i] = i;
int a, b;
scanf("%d%d", &a, &b);
vp.push_back(pr(a, b));
}
for(int i = 0; i < n; i++)
{
int xi = vp[i].first;
int yi = vp[i].second;
for(int j = i + 1; j < n; j++)
{
int xj = vp[j].first;
int yj = vp[j].second;
int d2 = abs(xi - xj)*abs(xi - xj) + abs(yi - yj)*abs(yi - yj);
if(d2 >= 10 * 10 && d2 <= 1000 * 1000)
{
ve.push_back(Edge(i, j, sqrt(d2)));
}
}
}
sort(ve.begin(), ve.end());
int cnt = 0;
double sum = 0;
for(int i = 0; i < ve.size(); i++)
{
int a = ve[i].from;
int b = ve[i].to;
double v = ve[i].val;
if(findset(a) != findset(b))
{
unionset(a, b);
cnt++;
sum += v * 100;
}
}
if(cnt == n - 1)
{
printf("%.1f\n", sum);
}
else
{
printf("oh!\n");
}
}
return 0;
}