SOJ 4482 忽悠大神(MST、点权映射到边权)

题意:

$N,M\le 10^5,N个点M条边无向图,点权W_i,边权C_i\le 1000$
$现要保证图联通的情况下删除最多的边$
$在此基础上,使得从某一起点出发,经过所有的点回到原点的权和最小$
$输出这个权和$

分析:

$我们发现因为要来回,所以边权其实是\times 2的,选了边其实2个端点必定会走$
$所以其实可以把点权映射到点上,即对于(u, v)新的边权为w_u+w_v+2\times c_{uv}$
$然后求一遍MST就好了,当然起点要再算一次,选择最小权的那个起点就好了$
$时间复杂度O(n+mlogm)$

代码:

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//
// Created by TaoSama on 2016-04-09
// Copyright (c) 2016 TaoSama. All rights reserved.
//
#pragma comment(linker, "/STACK:102400000,102400000")
#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;
}
} edge[N];
struct DSU {
int n, p[N];
void init(int _n) {
n = _n;
for(int i = 1; i <= n; ++i) p[i] = i;
}
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;
}
} dsu;
int w[N];
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);
while(scanf("%d%d", &n, &m) == 2) {
int ans = INF;
for(int i = 1; i <= n; ++i) {
scanf("%d", w + i);
ans = min(ans, w[i]);
}
for(int i = 1; i <= m; ++i) {
int u, v, c; scanf("%d%d%d", &u, &v, &c);
edge[i] = (Edge) {u, v, 2 * c + w[u] + w[v]};
}
sort(edge + 1, edge + 1 + m);
dsu.init(n);
int cnt = 0;
for(int i = 1; i <= m; ++i) {
int u = edge[i].u, v = edge[i].v;
if(dsu.unite(u, v)) {
ans += edge[i].c;
if(++cnt == n - 1) break;
}
}
printf("%d\n", ans);
}
return 0;
}


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