HDU 5624 KK's Reconstruction（暴力 + MST）

题意:

N≤2000个点,M≤15000条边,求所有MST中最大边−最小边的最小值,无MST输出−1

分析:

模拟MST生成的过程,根据Kruskal算法的正确性

从小往大加边,如果某条边形成环了,就删去环上的最小边把这条边丢进去

生成MST时就更新答案,尝试完所有边就是ans

时间复杂度为O(mlogm+nm)

代码:

//
//  Created by TaoSama on 2016-02-10
//  Copyright (c) 2016 TaoSama. All rights reserved.
//
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

int n, m;
struct Edge {
int u, v, c;
bool operator<(const Edge& e) const {
return c < e.c;
}
bool operator==(const Edge& e) const {
return u == e.u && v == e.v && c == e.c;
}
} edge[15005];

Edge minEdge;
vector<Edge> G[2005];

bool dfs(int u, int fa, int t, multiset<int>& s) {
if(u == t) return true;
for(Edge& e : G[u]) {
int v = e.v;
if(v == fa) continue;
if(dfs(v, u, t, s)) {
minEdge = min(minEdge, e);
return true;
}
}
return false;
}

void del(multiset<int>& s) {
int u = minEdge.u, v = minEdge.v, c = minEdge.c;
s.erase(s.find(c));
G[u].erase(find(G[u].begin(), G[u].end(), minEdge));
swap(minEdge.u, minEdge.v);
G[v].erase(find(G[v].begin(), G[v].end(), minEdge));

}

int p[2005];
int find(int x) {
return p[x] = p[x] == x ? x : find(p[x]);
}

bool unite(int u, int v) {
u = find(u), v = find(v);
if(u == v) return false;
p[u] = v;
return true;
}

int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);

int t; scanf("%d", &t);
while(t--) {
scanf("%d%d", &n, &m);
for(int i = 1; i <= m; ++i) {
int u, v, c; scanf("%d%d%d", &u, &v, &c);
edge[i] = (Edge) {u, v, c};
}
sort(edge + 1, edge + 1 + m);
multiset<int> s;
for(int i = 1; i <= n; ++i) {
p[i] = i;
G[i].clear();
}

int ans = 2e9, cnt = 0;
for(int i = 1; i <= m; ++i) {
int u = edge[i].u, v = edge[i].v, c = edge[i].c;
if(unite(u, v)) ++cnt;
else {
minEdge.c = 2e9;
dfs(u, -1, v, s);
del(s);
}
s.insert(c);
G[u].push_back((Edge) {u, v, c});
G[v].push_back((Edge) {v, u, c});
if(cnt == n - 1) ans = min(ans, c - *s.begin());
}
if(ans == 2e9) ans = -1;
printf("%d\n", ans);
}
return 0;
}

如果数据范围小的话,可以跑m遍Kruskal,易写,时间复杂度为O(mlogm+m2)

代码:

//
//  Created by TaoSama on 2016-02-06
//  Copyright (c) 2016 TaoSama. All rights reserved.
//
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

int n, m;
struct Edge {
int u, v, c;
void read() {scanf("%d%d%d", &u, &v, &c);}
bool operator<(const Edge& e) const {
return c < e.c;
}
} edge
;

int p
;
int find(int x) {
return p[x] = p[x] == x ? x : find(p[x]);
}

bool unite(int x, int y) {
x = find(x), y = find(y);
if(x == y) return false;
p[x] = y;
return true;
}

int kruskal(int i) {
if(m - i + 1 < n - 1) return 0x7fffffff;
int ret = -edge[i].c, cnt = 0, maxv = 0;
for(int j = 1; j <= n; ++j) p[j] = j;
for(int j = i; j <= m; ++j) {
int u = edge[j].u, v = edge[j].v;
cnt += unite(u, v);
if(cnt == n - 1) {
ret += edge[j].c;
break;
}
}
if(cnt != n - 1) return 0x7fffffff;
return ret;
}

int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);

int t; scanf("%d", &t);
while(t--) {
scanf("%d%d", &n, &m);
for(int i = 1; i <= m; ++i) edge[i].read();
sort(edge + 1, edge + 1 + m);
int ans = 0x7fffffff;
for(int i = 1; i <= m; ++i) {
ans = min(ans, kruskal(i));
}
if(ans == 0x7fffffff) ans = -1;
printf("%d\n", ans);
}
return 0;
}