HDU 1102 Constructing Roads

思想:简单最小生成树,kruskal。 结构体数组的范围一定要注意 不 要 开 小 了 ,因为这个问题时间超限n次,还找不出来o(╯□╰)o,还有就是注意一些细节问题。



#include
#include
#include
#include
#include
#include
using namespace std;
int per[105];
void init()
{
    for(int i = 0; i < 105; i ++)
    {
        per[i] = i;
    }
}
struct Edge
{
    int x;
    int y;
    int dis;
}edge[10005];//不要开小了;
bool cmp(Edge a, Edge b)
{
    return a.dis < b.dis;
}
int find(int x)
{
    int r = x;
    while(r != per[r])
        r = per[r];
    int i = x;
    int j;
    while(i != r)
    {
        j = per[i];
        per[i] = r;
        i = j;
    }
    return r;
}
void join(int x, int y)
{
    int s = find(x);
    int e = find(y);
    if(s != e)
    {
        per[s] = e;
    }
}
int kruskal(int n,int c)
{
    sort(edge,edge+c,cmp);
    int sum = 0,cnt = 0;
    for(int i = 0; i < c; i ++)
    {
        int s = find(edge[i].x);
        int e = find(edge[i].y);
        if(s != e)
        {
            join(s,e);
            sum += edge[i].dis;
            cnt++;
        }
        if(cnt == n-1)
            break;
    }
    return sum;
}
int main()
{
    int n,a,b;
    while(scanf("%d",&n)!= EOF)
    {
        init();
        int c = 0;
        for(int i = 1; i <= n; i ++)
        {
            for(int j = 1; j <= n; j ++)
            {
                scanf("%d",&a);
                edge[c].x = i;
                edge[c].y = j;
                edge[c++].dis = a;
            }
        }
        int q;
        scanf("%d",&q);
        for(int i = 0; i < q; i ++)
        {
            int a,b;
            scanf("%d%d",&a,&b);
            edge[(a-1)*n+b-1].dis = 0;
        }
        int d = kruskal(n,c);
        printf("%d\n",d);

    }


    return 0;
}


如有错误,欢迎指出~

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