SGU 323 Aviamachinations
Aviamachinations
This problem will be judged on SGU. Original ID: 323
64-bit integer IO format: %I64d Java class name: Solution
The Antimonopoly Committee was disbanded as a result of a government crisis. So, Don Berlione decided to close all but one airline. Naturally, this company should have flights (possibly including stopovers) from any town of Berland to any other one. To be able to choose the airline satisfying the above requirement, Don Berlione decided to carry out a number of fake purchase-sell operations. During a purchase-sell operation a flight of one airline is passed under the control of another airline. A purchase-sell operation is just a money transfer from one pocket to another. But still a special tax should be paid to the government for each operation.
So each flight is characterized by two towns it connects, the airline it belongs to and the tax amount that should be paid for a purchase-sell operation.
Your task is to find P — the minimum possible amount of money Don Berlione needs to spend to make it possible to leave only one airline carrying out flights (possibly with stopovers) from each town of Berland to any other. Also you need to suggest a plan of actions for Don Berlione.
Input
Output
Sample Input
sample input |
sample output |
4 3 4 |
5 2 1 |
#include <bits/stdc++.h>
using namespace std;
const int maxn = ;
class ARC {
public:
int u,v;
ARC(int x = ,int y = ) {
u = x;
v = y;
}
};
class ARC1:public ARC {
public:
int w,id;
ARC1(int x = ,int y = ,int cw = ,int cid = ):ARC(x,y) {
w = cw;
id = cid;
}
bool operator<(const ARC1 &t) {
return w < t.w;
}
} g[maxn];
class ARC2:public ARC {
public:
int next;
ARC2(int x = ,int y = ,int nxt = -):ARC(x,y) {
next = nxt;
}
} e[maxn];
int head[maxn],uf[maxn],tot,n,m,k;
void init() {
for(int i = ; i <= n; ++i) uf[i] = i;
}
int Find(int x) {
return uf[x] = x == uf[x]?x:Find(uf[x]);
}
void add(int u,int v,int x) {
e[tot] = ARC2(u,v,head[x]);
head[x] = tot++;
}
vector<int>MST,ans,tans;
void Kruskal() {
init();
MST.clear();
sort(g,g+k);
for(int i = ; i < k; ++i) {
int x = Find(g[i].u),y = Find(g[i].v);
if(x == y) continue;
uf[x] = y;
MST.push_back(i);
if(MST.size() >= n-) return;
}
}
void solve() {
int sum = 0x3f3f3f3f,id,tsum;
for(int i = ; i <= m; ++i) {
init();
tans.clear();
for(int j = head[i]; ~j; j = e[j].next) {
int x = Find(e[j].u),y = Find(e[j].v);
if(x != y) uf[x] = y;
}
for(int j = tsum = ; j < MST.size(); ++j) {
int x = Find(g[MST[j]].u),y = Find(g[MST[j]].v);
if(x == y) continue;
tsum += g[MST[j]].w;
uf[x] = y;
tans.push_back(MST[j]);
}
if(tsum < sum) {
sum = tsum;
id = i;
ans.clear();
std::copy(tans.begin(),tans.end(),std::back_inserter(ans));
}
}
printf("%d %d %d\n",sum,id,ans.size());
bool flag = false;
sort(ans.begin(),ans.end());
for(int i = ; i < ans.size(); ++i) {
if(flag) putchar(' ');
printf("%d",g[ans[i]].id);
flag = true;
}
putchar('\n');
}
int main() {
int a,b,c,p;
scanf("%d %d %d",&n,&m,&k);
memset(head,-,sizeof(head));
for(int i = tot = ; i < k; ++i) {
scanf("%d %d %d %d",&a,&b,&c,&p);
g[i] = ARC1(a,b,p,i + );
add(a,b,c);
}
Kruskal();
solve();
return ;
}
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