Uva 10765 点双连通求删任意点后剩下图中的连通分量数

标签：style   blog   os   io   2014   art   

题目挂在wustoj上，不知道什么原因不能粘贴链接。。

第一题题号为1314。。这题是智力题。。。换成7的阶乘就可以了。。

代码如下。

#include<cstdio>
int main()
{
   printf("...............................................................................................\n");
   printf("..#.........................#...........#.........................#............................\n");
   printf(".#...#####...#...........#...#.........#...#####...#...........#...#..............#####...#....\n");
   printf("#........#...#.....#.....#....#.......#........#...#...........#....#.................#...#....\n");
   printf("#.......#....#.....#.....#....#.......#.......#....#...........#....#....#####.......#....#....\n");
   printf("#......#.....#...#####...#....#...#...#......#.....#...#####...#....#...............#.....#....\n");
   printf("#.....#......#.....#.....#....#.......#.....#......#...........#....#....#####.....#......#....\n");
   printf("#.....#......#.....#.....#....#.......#.....#......#...........#....#..............#...........\n");
   printf(".#....#......#...........#....#........#....#......#...........#....#..............#......#....\n");
   printf("..#..........................#..........#..........................#...........................\n");
   printf("...............................................................................................\n");
   return 0;
}

for(i=2;i<=len;i++)
        {
           if(s[i]==s[i-1])
                a[i]++;
           a[i]=a[i]+a[i-1];
        }

完整代码为：

#include<cstdio>
#include<cstring>
int a[100002];
int main()
{
    char s[100002];
    int len,i,pos,j,m,le,ri;
    while(scanf("%s",s)!=EOF)
    {
        memset(a,0,sizeof(a));
        a[1]=0;
        len=strlen(s);
        for(i=len;i>=1;i--)
            s[i]=s[i-1];
        s[len+1]=‘\0‘;
        for(i=2;i<=len;i++)
        {
           if(s[i]==s[i-1])
             a[i]++;
           a[i]=a[i]+a[i-1];
        }
        scanf("%d",&m);
        for(i=1;i<=m;i++)
          {
              scanf("%d%d",&le,&ri);
              printf("%d\n",a[ri]-a[le]);
          }
    }
    return 0;
}

第四题为常数快速幂：

代码为：

#include<cstdio>
#define mode 100000007
int main()
{
    long long a,b,c;
    long long ans,muli;
    while(scanf("%lld%lld%lld",&a,&b,&c)!=EOF)
    {
        ans=c;
        muli=a;
        while(b)
        {
            if(b&1)//基数就乘
                ans=(ans*muli)%mode;
              muli=(muli*muli)%mode;
              b>>=1;
        }
        printf("%lld\n",ans%mode);
    }
    return 0;
}

#include<cstdio>
#include<cstring>
#define maxn 300
#define INF 0x3f3f3f3f
int cnt;
int a[maxn],edge[maxn][maxn],vis[maxn],b[maxn],dis[maxn];

int  dijkstra(int start,int End)
{
    int i,j,tmp,now;
    memset(vis,0,sizeof(vis));
    memset(dis,0x3f,sizeof(dis));
    dis[start]=0;
    for(i=1;i<=cnt;i++)
    {
        tmp=INF;
        for(j=1;j<=cnt;j++)
        {
            if(!vis[j]&&tmp>dis[j])
            {
                tmp=dis[j];
                now=j;
            }
        }
        vis[now]=1;
        for(j=1;j<=cnt;j++)
        {
            if(!vis[j]&&dis[j]>dis[now]+edge[now][j])
                dis[j]=dis[now]+edge[now][j];
         }
    }
    if(dis[End]<INF)
        return 1;
    return 0;
}

int main()
{
    int i,T,j;
    int flag,x,y;
    while(~scanf("%d",&T))
    {
        cnt=0;
        memset(edge,0x3f,sizeof(edge));
        for(i=1;i<=T;i++)
        {
            scanf("%d%d%d",&flag,&x,&y);
            if(flag==1)
            {
                cnt++;
                a[cnt]=x;
                b[cnt]=y;
                for(j=1;j<cnt;j++)
                {
                    if(x>a[j]&&x<b[j]||y>a[j]&&y<b[j]) edge[cnt][j]=1;
                    if(a[j]>x&&a[j]<y||b[j]>x&&b[j]<y) edge[j][cnt]=1;
                }
            }
            else
            {
                if(dijkstra(x,y))
                    printf("YES\n");
                else
                    printf("NO\n");
            }
        }
    }
    return 0;
}

代码如下：

#include<cstdio>
int root[10001];
int findroot(int x)
{
    if(root[x]!=x)
        root[x]=findroot(root[x]);
    return root[x];
}

int main()
{
    int T;
    int n,m,i,pos;
    int a,b,fx,fy;
    scanf("%d",&T);
    while(T--)
    {
        pos=0;
        scanf("%d%d",&n,&m);
        for(i=1;i<=n;i++)
            root[i]=i;
        for(i=1;i<=m;i++)
        {
            scanf("%d%d",&a,&b);
            fx=findroot(a);
            fy=findroot(b);
            if(fx!=fy)
                root[fx]=fy;
        }
        for(i=1;i<=n;i++)
        {
            if(root[i]==i)
                pos++;
        }
        printf("%d\n",pos);
    }
}

Uva 10765 点双连通求删任意点后剩下图中的连通分量数,码迷,mamicode.com
Uva 10765 点双连通求删任意点后剩下图中的连通分量数
标签：style   blog   os   io   2014   art   
原文地址：http://blog.csdn.net/acmmmm/article/details/24727629