zoukankan      html  css  js  c++  java
  • Conscription

    Conscription

    Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)
    Total Submission(s) : 13   Accepted Submission(s) : 6
    Problem Description

    Windy has a country, and he wants to build an army to protect his country. He has picked up N girls and M boys and wants to collect them to be his soldiers. To collect a soldier without any privilege, he must pay 10000 RMB. There are some relationships between girls and boys and Windy can use these relationships to reduce his cost. If girl x and boy y have a relationship d and one of them has been collected, Windy can collect the other one with 10000-d RMB. Now given all the relationships between girls and boys, your assignment is to find the least amount of money Windy has to pay. Notice that only one relationship can be used when collecting one soldier.

     

    Input

    The first line of input is the number of test case.
    The first line of each test case contains three integers, N, M and R.
    Then R lines followed, each contains three integers xi, yi and di.
    There is a blank line before each test case.

    1 ≤ N, M ≤ 10000
    0 ≤ R ≤ 50,000
    0 ≤ xi < N
    0 ≤ yi < M
    0 < di < 10000

     

    Output
    For each test case output the answer in a single line.
     

    Sample Input
    2 5 5 8 4 3 6831 1 3 4583 0 0 6592 0 1 3063 3 3 4975 1 3 2049 4 2 2104 2 2 781 5 5 10 2 4 9820 3 2 6236 3 1 8864 2 4 8326 2 0 5156 2 0 1463 4 1 2439 0 4 4373 3 4 8889 2 4 3133
     

    Sample Output
    71071 54223
    题解:
    征兵:有n个女生,m个男生,其中每个人需要花费10000元。不过男生和女生之间有相互作用,可以降低费用。比如编号为1的女生和编号为1的男生之间有关系d,那么女生已经征兵结束后,男生只需10000-d即可入伍
    kruskal算法,让男生直接+maxn就好;
    代码:
     1 #include<stdio.h>
     2 #include<string.h>
     3 #include<algorithm>
     4 using namespace std;
     5 const int MAXN=10010;
     6 struct Node{
     7     int s,e,c;
     8 };
     9 int cmp(Node a,Node b){
    10     return a.c>b.c;
    11 }
    12 Node dt[MAXN*5];
    13 int pre[MAXN*2];
    14 int ans;
    15 int find(int x){
    16     return pre[x]= x==pre[x]?x:find(pre[x]);
    17 }
    18 void initial(){
    19     memset(pre,-1,sizeof(pre));
    20     ans=0;
    21 }
    22 void merge(Node a){
    23         int f1,f2;
    24     if(pre[a.s]==-1)pre[a.s]=a.s;
    25     if(pre[a.e]==-1)pre[a.e]=a.e;
    26     f1=find(a.s);f2=find(a.e);
    27     if(f1!=f2){
    28         pre[f1]=f2;
    29         ans+=a.c;
    30     }
    31 }
    32 int main(){int N,M,R,T;
    33 scanf("%d",&T);
    34     while(T--){
    35         scanf("%d%d%d",&N,&M,&R);
    36         initial();
    37         for(int i=0;i<R;i++){
    38             scanf("%d%d%d",&dt[i].s,&dt[i].e,&dt[i].c);
    39             dt[i].e+=MAXN;
    40         }
    41         sort(dt,dt+R,cmp);
    42         for(int i=0;i<R;i++){
    43             merge(dt[i]);
    44         }
    45         printf("%d
    ",10000*(M+N)-ans);
    46     }
    47     return 0;
    48 }
  • 相关阅读:
    牛客 158F 青蛙 (贪心)
    牛客 158D a-贝利福斯数
    长沙理工大学第十二届ACM大赛-重现赛 大家一起来数二叉树吧 (组合计数)
    美团2017年CodeM大赛-初赛B轮 黑白树 (树形dp)
    美团2017年CodeM大赛-初赛A轮 C合并回文子串
    活动安排问题
    0和5
    1489 蜥蜴和地下室
    1067 Bash游戏 V2
    1062 序列中最大的数
  • 原文地址:https://www.cnblogs.com/handsomecui/p/4728499.html
Copyright © 2011-2022 走看看