题目
Bob was in trouble.He rubbed the magic ring on his finger, and you came out of the ground.
You are given an undirected graph G which contains n vertices labelled from 1 to n, with m weighted edges between them coloured in black or white.You have to choose some edges in G such that there is at least one path between any two vertices only passing by selected edges, and you can select no more than k white edges.There may be multiple available strategies to determine these edges, and you are asked to find out the way with a maximum total weight of edges.
Input
The first line contains an integer T (1≤T≤5) indicating the number of test cases.
For each test case, the first line contains three integers n (1≤n≤50000),m and k (1≤k≤m≤500000).
Each of the following m lines contains four integers u,v (1≤u,v≤n),w (0≤w≤100000) and c (0≤c≤1) describing an edge of weight w and colour c between the u-th vertex and the v-th vertex. Here an edge is white if c=1, or black if c=0.
Note that there may be multiple edges between some vertices, and self-loops are also allowed.
Output
For each test case, output a single line with an integer indicating the maximum total weight of selected edges, or output -1 if there is no solution for the given graph.
Sample Input
1
5 6 2
1 2 0 0
1 3 5 1
1 5 1 0
2 3 6 1
2 4 2 0
3 4 7 1
Sample Output
16
题意
给n个结点编号1~n，给m个有边权的边，边又分为白边与黑边，标记出其中若干个边，其中白边最多只能标记k条，令其中任何两个点都有一条路径只经过被标记的边，输出被标记的边权和，如果有多解，输出其中边权和最大的一种情况。
思路
因为黑边可以任意选，先选出所有黑边，并把其中黑边所连接的点标记出来，然后按边权从大到小遍历全部白边，如果能够选出k条白边令整个图联通就有解，反之无解。
PS.思路很简单，但是我码代码的能力实在太弱，本来一道水题代码写的又臭又长，好在一发a了，不然改代码能让我痛苦死…………
代码
#include<bits/stdc++.h>
using namespace std;
#define IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define ll long long
const int MAXN = 5e4 + 5, MAXM = 5e5 + 5;
int t, n, m, k;
struct node {
int u, v, w;
}white[MAXM], black[MAXM];
int wnum, bnum;
int vis[MAXN], jud[MAXN], cnt[MAXN];
bool cmp(node a, node b) {
return a.w > b.w;
}
int main(void){
IO;
cin >> t;
while(t--) {
memset(vis,0,sizeof(vis));
memset(jud,0,sizeof(jud));
memset(cnt,0,sizeof(cnt));
memset(white,0,sizeof(white));
memset(black,0,sizeof(black));
wnum = 0, bnum = 0;
cin >> n >> m >> k;
for(int i = 0; i < m; i++) {
int c;node x;cin >> x.u >> x.v >> x.w >> c;
if(c)white[++wnum] = x;
else black[++bnum] = x;
}
jud[1] = 1;int num = 0, now = 1;
ll ans = 0;
for(int i = 1; i <= bnum; i++) {
if(!vis[black[i].u])vis[black[i].u] = 1,num++;
if(!vis[black[i].v])vis[black[i].v] = 1,num++;
ans += black[i].w;
}
sort(white + 1, white + wnum + 1, &cmp);
queue<int> q;
for(int i = 1; i <= wnum; i++) {
if(!vis[white[i].u] && !vis[white[i].v]) {
ans += white[i].w;
k--;
now++;
vis[white[i].u] = now;
vis[white[i].v] = now;
cnt[now] += 2;
}else if(!vis[white[i].u]) {
ans += white[i].w;
k--;
vis[white[i].u] = vis[white[i].v];
cnt[vis[white[i].v]]++;
if(jud[vis[white[i].v]])num++;
}else if(!vis[white[i].v]) {
ans += white[i].w;
k--;
vis[white[i].v] = vis[white[i].u];
cnt[vis[white[i].u]]++;
if(jud[vis[white[i].u]])num++;
}else if(vis[white[i].u] == vis[white[i].v]) {
q.push(white[i].w);
}else if(vis[white[i].u] != vis[white[i].v]) {
if(jud[vis[white[i].u]] && jud[vis[white[i].v]]) {
q.push(white[i].w);
}else if(jud[vis[white[i].u]] || jud[vis[white[i].v]]) {
ans += white[i].w;
k--;
if(jud[vis[white[i].u]])num += jud[vis[white[i].v]];
else num += jud[vis[white[i].u]];
jud[vis[white[i].u]] = jud[vis[white[i].v]] = 1;
}else {
int mi = min(vis[white[i].u], vis[white[i].v]), ma = max(vis[white[i].u], vis[white[i].v]);
for(int i = 1; i <= n; i++) {
if(vis[i] == mi)vis[i] == ma;
}
cnt[ma] += cnt[mi];
cnt[mi] = 0;
}
}
if(num == n) {
if(k <= 0)break;
while(!q.empty()) {
ans += q.front();
q.pop();
k--;
if(k == 0)break;
}
for(int j = i + 1; j <= wnum && k > 0; j++) {
ans += white[j].w;
k--;
}
break;
}
}
if(num == n) {
if(k >= 0) {
cout << ans << '\n';
}else {
cout << -1 << '\n';
}
}else {
cout << -1 << '\n';
}
}
}