kuangbin 专题八 生成树

POJ-1679 The Unique MST

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#include <cstdio>
#include <algorithm>
#define N 101
using namespace std;
int t, n, m, p[N];
struct edge
{
int w, u, v;
bool operator<(const edge &a) const { return w < a.w; }
} e[N * N];
int find(int x)
{
return p[x] == x ? x : p[x] = find(p[x]);
}
void kruskal()
{
int sum = 0, res = 0;
for (int i = 1; i <= n; ++i)
p[i] = i;
sort(e, e + m);
for (int i = 0; i < m; ++i)
{
int r = i;
while (r < m && e[i].w == e[r].w)
++r;
--r;
for (int j = i; j <= r; ++j)
{
int a = find(e[j].u), b = find(e[j].v);

if (a != b)
sum += e[j].w;
}
for (int j = i; j <= r; ++j)
{
int a = find(e[j].u), b = find(e[j].v);
if (a != b)
p[a] = b, res += e[j].w;
}
i = r;
}
if (sum != res)
printf("Not Unique!\n");
else
printf("%d\n", res);
}
int main()
{
scanf("%d", &t);
while (t--)
{
scanf("%d%d", &n, &m);
for (int i = 0; i < m; ++i)
scanf("%d%d%d", &e[i].u, &e[i].v, &e[i].w);
kruskal();
}
return 0;
}

HDU-4081 Qin Shi Huang’s National Road System

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#include<bits/stdc++.h>
#define N 1001
#define M 2002
using namespace std;
int T, n, x[N], y[N], p[N];
bool st[N];
double g[N][N], d[N], w[M];
int h[N], e[M], ne[M], tot;
int path[N];
vector<pair<int, int>> vp;
void add(int u, int v, int ww)
{
e[tot] = v, w[tot] = ww, ne[tot] = h[u], h[u] = tot++;
}
double prim()
{
double res = 0;
for (int i = 0; i < n; ++i)
d[i] = 0x3f3f3f3f, st[i] = 0;
d[0] = 0;
for (int i = 0; i < n; ++i)
{
int t = -1;
for (int j = 0; j < n; ++j)
if (!st[j] && (t == -1 || d[j] < d[t]))
t = j;
st[t] = 1;
if (i)
{
res += d[t];
add(path[t], t, g[path[t]][t]);
add(t, path[t], g[path[t]][t]);
vp.push_back({t, path[t]});
}
for (int j = 0; j < n; ++j)
if (!st[j] && d[j] > g[t][j])
d[j] = g[t][j], path[j] = t;
}
return res;
}
int main()
{
scanf("%d", &T);
while (T--)
{
scanf("%d", &n);
memset(h, -1, sizeof h);
tot = 0;
vp.clear();
for (int i = 0; i < n; ++i)
scanf("%d%d%d", x + i, y + i, p + i);
for (int i = 0; i < n; ++i)
{
g[i][i] = 0;
for (int j = i + 1; j < n; ++j)
g[i][j] = g[j][i] = sqrt(pow(x[i] - x[j], 2) + pow(y[i] - y[j], 2));
}
double mst = prim(), ans = 0;
for (auto &[u, v] : vp)
{
int p1 = 0, p2 = 0;
function<void(int, int, int&)> dfs = [&](int f, int u, int& pv)
{
pv = max(pv, p[u]);
for (int i = h[u]; ~i; i = ne[i])
if (e[i] != f)
dfs(u, e[i], pv);
};
dfs(u, v, p1), dfs(v, u, p2);
ans = max(ans, (p1 + p2) / (mst - g[u][v]));
}
printf("%.2lf\n", ans);
}
return 0;
}

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HDU-4009 Transfer water

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SPOJ-DETER3 Find The Determinant III

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HDU-4305 Lightning

HDU-4408 Minimum Spanning Tree

SPOJ-HIGH Highways