Conscription
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
翻译一下就是ｎ个女孩，ｍ个男孩，全部要招募，但是有些女孩ｘ和男孩ｙ有关系，如果已经招募了其中一个，另一个就只需要10000- ｄ了,求最小花费。
开始是想最小生成树，求招募ｍ＋ｎ个人最小花费，预处理时用10000先减去d，再求最小生成树，但是图可能不是连通图，所以对每个联通子图求最小生成树，再加上树的个数 num*10000，当然，一个人也是一棵树，可就是一直ｗａ。
但是，如果反过来，求ｎ＋ｍ个人能优惠的钱数，最后用总钱数再减去它就行了，也不用预处理。
#include<iostream>
#include<algorithm>
using namespace std;
char buf[100000],*L=buf,*R=buf;
#define gc() L==R&&(R=(L=buf)+fread(buf,1,100000,stdin),L==R)?EOF:*L++;
template<typename T>
inline void read(T&x) {
char ch=gc();
x=0;
while (ch<'0'||ch>'9')
ch=gc();
while (ch>='0'&&ch<='9')
x=(x<<3)+(x<<1)+ch-48,ch=gc();
}
const int MAXN=5e4+20;
int fa[MAXN],Rank[MAXN];
inline int find(int x) {
while(fa[x]!=x)
x=fa[x]=fa[fa[x]];
return x;
}
inline void merge(int x,int y) {
int xx=find(x),yy=find(y);
if(xx==yy)return;
if(Rank[xx]<Rank[yy])fa[xx]=yy;
else {
fa[yy]=xx;
if(Rank[xx]==Rank[yy])++Rank[xx];
}
}
inline void init() {
for(int i=0; i<MAXN; ++i){
fa[i]=i,Rank[i]=0;
}
}
struct Node {
int x,y,v;
bool operator<(const Node&b)const {
return v>b.v;
}
} p[MAXN];
int main() {
freopen("in.txt","r",stdin);
int t;
read(t);
while(t--) {
int n,m,r;
read(n),read(m),read(r);
init();
for(int i=0; i<r; ++i) {
read(p[i].x),read(p[i].y),read(p[i].v);
p[i].y+=n;//女生: [0,n-1],男生: [n,m+n];
}
int ans=0;//优惠多少钱
sort(p,p+r);
for(int i=0; i<r; ++i) {
if(find(p[i].x)==find(p[i].y))
continue;
merge(p[i].x,p[i].y);
ans+=p[i].v;
//加个边数判断也行
}
printf("%d
",(m+n)*10000-ans);//总钱数减去优惠的钱数
}
return 0;
}
//正着算，居然过了
#include<iostream>
#include<algorithm>
using namespace std;
char buf[100000],*L=buf,*R=buf;
#define gc() L==R&&(R=(L=buf)+fread(buf,1,100000,stdin),L==R)?EOF:*L++;
template<typename T>
inline void read(T&x) {
char ch=gc();
x=0;
while (ch<'0'||ch>'9')
ch=gc();
while (ch>='0'&&ch<='9')
x=(x<<3)+(x<<1)+ch-48,ch=gc();
}
const int MAXN=5e4+20;
int fa[MAXN],Rank[MAXN];
inline int find(int x) {
while(fa[x]!=x)
x=fa[x]=fa[fa[x]];
return x;
}
inline void merge(int x,int y) {
int xx=find(x),yy=find(y);
if(xx==yy)return;
if(Rank[xx]<Rank[yy])fa[xx]=yy;
else {
fa[yy]=xx;
if(Rank[xx]==Rank[yy])++Rank[xx];
}
}
inline void init() {
for(int i=0; i<MAXN; ++i){
fa[i]=i,Rank[i]=0;
}
}
struct Node {
int x,y,v;
bool operator<(const Node&b)const {
return v<b.v;
}
} p[MAXN];
int main() {
freopen("in.txt","r",stdin);
int t;
read(t);
while(t--) {
int n,m,r;
read(n),read(m),read(r);
init();
for(int i=0; i<r; ++i) {
read(p[i].x),read(p[i].y),read(p[i].v);
p[i].y+=n;
p[i].v=10000-p[i].v;
}
int ans=0;
sort(p,p+r);
for(int i=0; i<r; ++i) {
if(find(p[i].x)==find(p[i].y))
continue;
merge(p[i].x,p[i].y);
ans+=p[i].v;
}
int num=0;
for(int i=0;i<m+n;++i)
if(fa[i]==i)++num;
printf("%d
",ans+num*10000);
}
return 0;
}