HDU 4253 Two Famous Companies
Two Famous Companies
This problem will be judged on HDU. Original ID: 4253
64-bit integer IO format: %I64d Java class name: Main
Input
Output
Sample Input
2 2 1
0 1 1 1
0 1 2 0
2 2 0
0 1 1 1
0 1 2 0
Sample Output
Case 1: 2
Case 2: 1
In the first case, there are two cable plans between the only two cities, one from China Telecom and one from China Unicom. Mr. B needs to choose the one from China Telecom to satisfy the problem requirement even the cost is higher. In the second case, Mr. B must choose the cable from China Unicom, which leads the answer to 1.
Source
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <vector>
#include <queue>
#include <cstdlib>
#include <string>
#include <set>
#include <stack>
#define LL long long
#define pii pair<int,int>
#define INF 0x3f3f3f3f
using namespace std;
const int maxn = ;
struct arc{
int u,v,w,id;
arc(int uu = ,int vv = ,int ww = ,int iid = ){
u = uu;
v = vv;
w = ww;
id = iid;
}
bool operator<(const arc &tmp) const{
return w < tmp.w;
}
};
arc e[][maxn];
int uf[maxn],n,m,k,tot1,tot2,cost;
int Find(int x){
if(x == uf[x]) return x;
return uf[x] = Find(uf[x]);
}
bool uset(int u,int v){
int tx = Find(u);
int ty = Find(v);
if(tx != ty) uf[tx] = ty;
return tx != ty;
}
bool check(int delta){
int i,j,cnt;
for(i = ; i <= n; ++i) uf[i] = i;
i = j = cnt = cost = ;
arc tmp;
while(i < tot1 || j < tot2){
if(e[][i].w + delta <= e[][j].w){
tmp = e[][i++];
tmp.w += delta;
}else tmp = e[][j++];
if(uset(tmp.u,tmp.v)){
cost += tmp.w;
if(!tmp.id) cnt++;
}
}
return cnt >= k;
}
int main() {
int u,v,w,id,cs = ;
while(~scanf("%d %d %d",&n,&m,&k)){
for(int i = tot1 = tot2 = ; i < m; ++i){
scanf("%d %d %d %d",&u,&v,&w,&id);
if(id) e[][tot2++] = arc(u,v,w,id);
else e[][tot1++] = arc(u,v,w,id);
}
e[][tot1].w = e[][tot2].w = INF;
sort(e[],e[]+tot1);
sort(e[],e[]+tot2);
int low = -,high = ,mid,delta;
while(low <= high){
mid = (low + high)>>;
if(check(mid)){
delta = mid;
low = mid + ;
}else high = mid - ;
}
check(delta);
printf("Case %d: %d\n",cs++,cost - delta*k);
}
return ;
}
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