题目意思:给出n个点,m条边,边分为两种,一种是A公司的,一种是B公司的。边上有权值,
问用n-1条边把n个点连起来的最小费用是多少,其中A公司的边刚好有k条。题目保证有解。
题解:题目意思很简单就是求MST且A公司要有且仅有k条边在树中,刚开始做的时候用贪心的方式求
最小生成树结果WA,后来看了下别人的题解,二分出一个最大值delta使得A公司的边加上这个值后再
求MST时A公司的边有大于等于k条,然后答案就是cost of MST - k * delta。思想就是用一个delta去
逼近答案。当delta越大的时候A公司的边就越少,反之越多。所以二分delta可以找到使得A公司刚好取K条边的要求。
#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
struct node
{int u,v,cost;bool operator < (node a)const{return cost < a.cost;}
}arr[2][100005];
int father[100005];
int n,m,k,num[2],mincost;
int find(int x){if( x != father[x])return father[x] = find(father[x]);return father[x];
}
int union_set(int x,int y){x = find(x);y = find(y);if( x != y){father[x] = y;return 1;}return 0;
}
bool check(int delta){int tel = k,Union = n-k-1;int nn = n,i,j;mincost = 0;for( i = 0; i <= n; i++)father[i] = i;i = j = 0;while( nn > 1 ){if( j < num[1] && ( i >= num[0] || arr[0][i].cost + delta > arr[1][j].cost)){if( union_set(arr[1][j].u,arr[1][j].v ) ){mincost += arr[1][j].cost;nn--;Union--;}j++;}else{if( union_set(arr[0][i].u,arr[0][i].v ) ){mincost += arr[0][i].cost + delta;nn--;tel --;}i++;}}return tel <= 0;
}
int main(){struct node tmp;int cas = 1;int u,v,cost,x;int l,r,mid,m;freopen("in.txt","r",stdin);while( ~scanf("%d%d%d",&n,&m,&k)){num[0] = num[1] = r = 0;for(int i = 0; i < m; i++){father[i] = i;scanf("%d%d%d%d",&tmp.u,&tmp.v,&tmp.cost,&x);arr[x][num[x]++] = tmp;}for( int i = 0; i < 2; i++)sort(arr[i],arr[i]+num[i]);l = -110,r = 110;int res;while( l <= r){mid = (l+r)>>1;if( check(mid) ){res = mincost;m = mid;l = mid + 1;}else r = mid - 1;}printf("Case %d: %d\n",cas++,res - m*k);}
}