poj 3723 Conscription

好忧伤，根本就是同一种算法，为毛我的代码不能过啊！！！

就是kruskal的最大生成树变形。

Conscription

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 7493 Accepted: 2563

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 x_i, y_i and d_i.
There is a blank line before each test case.

1 ≤ N, M ≤ 10000
0 ≤ R ≤ 50,000
0 ≤ x_i < N
0 ≤ y_i < M
0 < d_i < 10000

Output

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

Sample Input

25 5 84 3 68311 3 45830 0 65920 1 30633 3 49751 3 20494 2 21042 2 7815 5 102 4 98203 2 62363 1 88642 4 83262 0 51562 0 14634 1 24390 4 43733 4 88892 4 3133

Sample Output

7107154223

Source

POJ Monthly Contest – 2009.04.05, windy7926778

贴下代码啦：

我的WA了：

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>using namespace std;const int maxn=20000;int father[maxn+10],rank[maxn+10];void init(){    for(int i=0;i<maxn+10;i++)    {        father[i]=i;        rank[i]=0;    }}int find(int x){    if(father[x]==x)        return x;    else        return father[x]=find(father[x]);}void unite(int x,int y){    x=find(x);    y=find(y);    if(x==y)        return;    if(rank[x]<rank[y])    {        father[x]=y;    }    else    {        father[y]=x;        if(rank[x]==rank[y])            rank[x]++;    }}struct edge{    int from;    int to;    int cost;};edge G[50000+10];bool cmp(const edge & a,const edge & b){    return a.cost>b.cost;}int t;int main(){    //freopen("x.in","r",stdin);    cin>>t;    while(t--)    {        init();        int n,m,r;        scanf("%d%d",&n,&m);        scanf("%d",&r);        for(int i=0;i<r;i++)        {            scanf("%d%d%d",&G[i].from,&G[i].to,&G[i].cost);            G[i].to+=m;        }        sort(G,G+r,cmp);        int sum=0;        for(int i=0;i<r;i++)        {            if(find(G[i].from)!=find(G[i].to))            {            unite(G[i].from,G[i].to);            sum+=G[i].cost;            }        }        printf("%d\n",10000*(n+m)-sum);    }    return 0;}

别人的。。。AC了：

#include <cstdio>#include <algorithm>#define N 20005#define E 50005using namespace std;struct edge{    int u,v,w;}e[E];int n,sum;int father[N],rank[N];bool cmp(edge a,edge b){    return a.w>b.w;}void makeset(void){    register int i=0;    for(i=0;i<n;i++)    {        father[i]=i;        rank[i]=0;    }}int find(int x){    if(x!=father[x])    father[x]=find(father[x]);    return father[x];}void Union(int x,int y,int w){    x=find(x),y=find(y);    if(x==y)    return;    if(rank[x]>rank[y])  father[y]=x;    else    {        if(rank[x]==rank[y])    rank[y]++;        father[x]=y;    }    sum+=w;}int main(void){    int m,r,t,a,b,c;    register int i;    scanf("%d",&t);    while(t--)    {        scanf("%d%d%d",&m,&n,&r);        n+=m;        makeset();        for(i=0;i<r;i++)        {            scanf("%d%d%d",&a,&b,&c);            b+=m;            e[i].u=a,e[i].v=b,e[i].w=c;        }        sort(e,e+r,cmp);        for(sum=i=0;i<r;i++)            Union(e[i].u,e[i].v,e[i].w);        printf("%d\n",10000*n-sum);    }    return 0;}

~~~~(>_<)~~~~ ，哪位大神告诉我为毛错了啊。。。