poj 3723 Conscription

最新推荐文章于 2019-07-25 21:06:53 发布

尘飞acmer

最新推荐文章于 2019-07-25 21:06:53 发布

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分类专栏： acm初级训练文章标签： acm kruskal 图论

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本文深入探讨了kruskal算法的变形在解决特定问题中的应用，具体涉及到Windy建立军队保护国家的问题。通过分析输入案例，解释了如何通过关系网络降低收集士兵的成本，并提供了两种不同的代码实现方式。尽管存在一些实现细节上的差异，但最终目标都是为了寻找最优化的解决方案，即最小化所需支付的费用。

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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: 65536K
Total 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

Sample Output

71071
54223

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 50005
using 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。。。