洛谷_Cx的故事_解题报告_第四题70

1.并查集求最小生成树

Code：

 #include <stdio.h>
 #include <stdlib.h>
  
 struct node
 {
     long x,y,c;
 }road[];
  
 long fa[];
  
 int cmp(const void *a,const void *b)
 {
     if ((*(struct node *)a).c < (*(struct node *)b).c)
         return -;
     else
         return ;
 }
  
 long getfather(long d)
 {
     if (fa[d]==d)
         return d;
     fa[d]=getfather(fa[d]);
     return fa[d];
 }
  
 int main()
 {
     long n,m,t,xx,yy,x,y,i,cnt;
     long long cost=;
     scanf("%ld%ld",&n,&m);
     cnt=n-;
     for (i=;i<=n;i++)
         fa[i]=i;
     for (i=;i<=m;i++)
     {
         scanf("%ld%ld",&x,&y);
         xx=getfather(x);
         yy=getfather(y);
         if (xx!=yy)
         {
             fa[xx]=yy;
             cnt--;
             if (cnt==)
             {
                 printf("0\n");
                 return ;
             }
         }
  
     }
     scanf("%ld",&t);
     for (i=;i<t;i++)
         scanf("%ld%ld%ld",&road[i].x,&road[i].y,&road[i].c);
     qsort(road,t,sizeof(struct node),cmp);
     for (i=;i<t;i++)
     {
         xx=getfather(road[i].x);
         yy=getfather(road[i].y);
         if (xx!=yy)
         {
             fa[xx]=yy;
             cost+=road[i].c;
             cnt--;
             if (cnt==)
             {
                 printf("%lld\n",cost);
                 return ;
             }
         }
     }
     printf("-1\n");
     return ;
 }

2.点有权值，spfa

 #include <stdio.h>
 #include <stdlib.h>
 #include <stdbool.h>
  
 struct node
 {
     long c;
     struct node *next;
 }*city[];
  
 long q[],pre[],m;
  
 long min(long a,long b)
 {
     if (a>b)
         return b;
     else
         return a;
 }
  
 void print(long d)
 {
     if (d!=m)
         print(pre[d]);
     printf("%ld",d);
     if (d!=)
         printf("->");
     else
         printf("\n");
 }
  
 int main()
 {
     long i,j,s,t,b,e,n,ren,g[],head,tail,sum[];
     bool vis[];
     struct node *p;
     scanf("%ld%ld%ld",&n,&m,&ren);
     for (i=;i<=n;i++)
     {
         scanf("%ld",&g[i]);
         scanf("%ld",&s);
         for (j=;j<=s;j++)
         {
             scanf("%ld",&t);
             p=(struct node *) malloc (sizeof(struct node));
             p->c=t;
             p->next=city[i];
             city[i]=p;
         }
     }
     for (i=;i<=n;i++)
     {
         vis[i]=true;
         sum[i]=;
     }
     sum[m]=g[m];
     head=;
     tail=;
     q[]=m;
  
     do
     {
         head=(head+)%;
 //        head++;
         b=q[head];
         p=city[b];
         while (p)
         {
             e=p->c;
             if (sum[b]+g[e]<sum[e])
             {
                 sum[e]=sum[b]+g[e];
                 pre[e]=b;
                 if (vis[e])
                 {
                     tail=(tail+)%;
 //                    tail++;
                     q[tail]=e;
                     vis[e]=false;
                 }
             }
             p=p->next;
         }
         vis[b]=true;
     }
     while (head!=tail);
  
     if (sum[]<=ren*)
     {
         print();
         printf("%ld\n",ren-sum[]/);
     }
     else
         printf("No way!");
     return ;
 }

3.

地图bfs

U10278 Cx的金字塔 _ 落谷1126机器人搬重物 解题报告

I.哪个是行，哪个是列，行列从哪边开始

II.一次操作

1.前移1,2,3！步

2.转90度

III.边界范围 1~n-1 1~m-1

IV.起始点不可用

V.起始点=终止点

Code：

 #include <stdio.h>
 #include <stdlib.h>
 #include <stdbool.h>
  
 struct node
 {
     long x,y,c,step;
 }q[];
  
 bool vis[][][];
  
 long head,tail,s,t;
  
 void add(long x,long y,long c)
 {
     tail++;
     q[tail].x=x;
     q[tail].y=y;
     q[tail].c=c;
     q[tail].step=q[head].step+;
     vis[x][y][c]=false;
     if (x==s && y==t)
     {
         printf("%ld\n",q[tail].step);
         exit();
     }
 }
  
 int main()
 {
     long n,m,i,j,k,x,y,c;
     long map[][];
     char ccs,cs;
     scanf("%ld%ld",&n,&m);
     for (i=;i<=n;i++)
         for (j=;j<=m;j++)
             for (k=;k<;k++)
                 vis[i][j][k]=true;
  
     for (i=;i<=n;i++)
         for (j=;j<=m;j++)
             scanf("%ld",&map[i][j]);
     scanf("%ld%ld%ld%ld%c%c",&x,&y,&s,&t,&ccs,&cs);
  
     if (x== || x==n || y== || y==m || map[x][y]== || map[x+][y]== || map[x][y+]== || map[x+][y+]==)
     {
         printf("-1");
         return ;
     }
  
     if (x==s && y==t)
     {
         printf("0\n");
         return ;
     }
  
     head=;
     tail=;
     q[].x=x;
     q[].y=y;
     if (cs=='E')
         c=;
     else if (cs=='S')
         c=;
     else if (cs=='W')
         c=;
     else
         c=;
     q[].c=c;
     q[].step=;
     vis[x][y][c]=false;
     //上：0 下：1 左：2 右：3
     head=;
     tail=;
     do
     {
         head++;
         x=q[head].x;
         y=q[head].y;
         c=q[head].c;
         //上
         if (c==)
         {
             if (x> && map[x-][y]== && map[x-][y+]==)
             {
                 if (vis[x-][y][c])
                     add(x-,y,c);
                 if (x> && map[x-][y]== && map[x-][y+]==)
                 {
                     if (vis[x-][y][c])
                         add(x-,y,c);
                     if (x> && map[x-][y]== && map[x-][y+]== && vis[x-][y][c])
                         add(x-,y,c);
                 }
             }
         }
         //下
         else if (c==)
         {
             if (x<n- && map[x+][y]== && map[x+][y+]==)
             {
                 if (vis[x+][y][c])
                     add(x+,y,c);
                 if (x<n- && map[x+][y]== && map[x+][y+]==)
                 {
                     if (vis[x+][y][c])
                         add(x+,y,c);
                     if (x<n- && map[x+][y]== && map[x+][y+]== && vis[x+][y][c])
                         add(x+,y,c);
                 }
             }
  
         }
         //左
         else if (c==)
         {
             if (y> && map[x][y-]== && map[x+][y-]==)
             {
                 if (vis[x][y-][c])
                     add(x,y-,c);
                 if (y> && map[x][y-]== && map[x+][y-]==)
                 {
                     if (vis[x][y-][c])
                         add(x,y-,c);
                     if (y> && map[x][y-]== && map[x+][y-]== && vis[x][y-][c])
                         add(x,y-,c);
                 }
             }
         }
         //右
         else
         {
             if (y<m- && map[x][y+]== && map[x+][y+]==)
             {
                 if (vis[x][y+][c])
                     add(x,y+,c);
                 if (y<m- && map[x][y+]== && map[x+][y+]==)
                 {
                     if (vis[x][y+][c])
                         add(x,y+,c);
                     if (y<m- && map[x][y+]== && map[x+][y+]== && vis[x][y+][c])
                         add(x,y+,c);
                 }
             }
         }
         if (c== || c==)
         {
             if (vis[x][y][])
                 add(x,y,);
             if (vis[x][y][])
                 add(x,y,);
         }
         else
         {
             if (vis[x][y][])
                 add(x,y,);
             if (vis[x][y][])
                 add(x,y,);
         }
     }
     while (head<tail);
     printf("-1\n");
  
     return ;
 }

4.

对于70%的数据：

2<=a<=250

对于100%的数据：

2<=a<=2^60

对于70%的数据：

初始：

现有一个花坛（A坛）装满了白蛇根草，还有两个空花坛（B坛，C坛），

三个花坛共有250*250*250=15625000种

(x,y,z)->其它状态

当有其中两个花坛的数目为|A|/2时结束

by lzu_cgb
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