应用了并查集判环的最小生成树算法
 #include <stdio.h>
#include <stdlib.h>
const int LEN = 1005;
int p[LEN];
void init ( int n )
   {
for ( int i=0; i<n; i++ )
  {
p[i] = -1;
 }
 }
int find ( int a )
{
if ( p[a] < 0 )
{
return a;
}
int r = find ( p[a] );
p[a] = r;
return r;
}
void un ( int a, int b )
{
int ra = find ( a );
int rb = find ( b );
if ( p[ra] < p[rb] )
{
p[ra] += p[rb];
p[rb] = ra;
}
else
{
p[rb] += p[ra];
p[ra] = rb;
}
}
struct
{
int b;
int e;
int len;
}seg[LEN*20];
int cmp ( const void *a, const void *b )
{
return *( ( int * )b + 2 ) - *( ( int * )a + 2 );
}
int kul ( int n, int m )
{
qsort ( seg, m, sizeof ( seg[0] ), cmp );
init ( n );
int sum = 0;
for ( int i=0; i<m; i++ )
{
if ( find ( seg[i].b ) != find ( seg[i].e ) )
{
sum += seg[i].len;
un ( seg[i].b, seg[i].e );
}
}
for ( i=0; i<n-1; i++ )
{
if ( find ( i ) != find ( i+1 ) )
{
return -1;
}
}
return sum;
}
int main ()
{
int n, m;
int b, e, len;
while ( scanf ( "%d%d", &n, &m ) != EOF )
{
for ( int i=0; i<m; i++ )
{
scanf ( "%d%d%d", &b, &e, &len );
seg[i].b = b - 1;
seg[i].e = e - 1;
seg[i].len = len;
}
printf ( "%d\n", kul ( n, m ) );
}
return 0;
}
简单的快排,傻傻的用了自己写的模版,呵呵
#include "stdio.h"
int num[100005];
void swap(int a,int b)
{
int t;
t=num[a];
num[a]=num[b];
num[b]=t;
}
int partion(int low,int high,int p)
{
while(low<=high)
{
if(low<p)
{
if(num[low]>num[p]){swap(low,p);p=low;}
low++;
}
else
{
if(high>p)
{
if(num[high]<num[p]){swap(high,p);p=high;}
}
high--;
}
}
return p;
}
void qsort(int left,int right)
{
int middle;
if(left<right)
{
middle=(left+right)/2;
middle=partion(left,right,middle);
qsort(left,middle-1);
qsort(middle+1,right);
}
}
int main()
{
int n,k,i,q;
char temp[5];
while(scanf("%d",&n)!=EOF)
{
for(i=0;i<n;i++)
{
scanf("%d",&num[i]);
}
scanf("%s",temp);
qsort(0,n-1);
scanf("%d",&k);
for(i=0;i<k;i++)
{
scanf("%d",&q);
printf("%d\n",num[q-1]);
}
}
return 0;
}
摘要: DFS搜索,是stick的简化版本
#include <stdio.h>#include <stdlib.h>int stick[64];int used[64];int n, m, len;void init ( int n ){ ... 阅读全文
摘要: 线段树和树状数组都可以用
#include <stdio.h>#include <stdlib.h>const int LEN = 15005;//坐标处理struct NODE{ int x; //X坐标 &nbs... 阅读全文
动态规划
#include <stdio.h>
const int LEN = 101;
char hash[26];
int r, c, n;
char map[LEN][LEN];
char cmap[LEN][LEN];
void cpy ()
{
for ( int i=0; i<r; i++ )
{
for ( int j=0; j<c; j++ )
{
map[i][j] = cmap[i][j];
}
}
}
int move[4][2] = { 0, 1, 1, 0, 0, -1, -1, 0 };
void ser ()
{
hash[ 'R'-'A' ] = 'S';
hash[ 'S'-'A' ] = 'P';
hash[ 'P'-'A' ] = 'R';
int flag = 1;
for ( int i=0; i<n; i++ )
{
flag = 0;
for ( int j=0; j<r; j++ )
{
for ( int z=0; z<c; z++ )
{
int tr, tc;
for ( int x =0; x<4; x++ )
{
tr = j+move[x][0];
tc = z+move[x][1];
if ( tr>=0 && tr<r && tc>=0 && tc<c )
{
if ( hash[ map[tr][tc]-'A' ] == map[j][z] )
{
cmap[j][z] = map[tr][tc];
flag = 1;
break;
}
}
}
if ( x >= 4 )
{
cmap[j][z] = map[j][z];
}
}
}
cpy ();
if ( ! flag )
{
break;
}
}
}
int main ()
{
int t;
scanf ( "%d", &t );
while ( t -- )
{
scanf ( "%d%d%d", &r, &c, &n );
getchar ();
for ( int i=0; i<r; i++ )
{
for ( int j=0; j<c; j++ )
{
scanf ( "%c", &map[i][j] );
}
getchar ();
}
ser ();
for ( i=0; i<r; i++ )
{
for ( int j=0; j<c; j++ )
{
printf ( "%c", map[i][j] );
}
printf ( "\n" );
}
printf ( "\n" );
}
return 0;
}
广搜题目
#include <stdio.h>
const int LEN = 101;
char hash[26];
int r, c, n;
char map[LEN][LEN];
char cmap[LEN][LEN];
void cpy ()
{
for ( int i=0; i<r; i++ )
{
for ( int j=0; j<c; j++ )
{
map[i][j] = cmap[i][j];
}
}
}
int move[4][2] = { 0, 1, 1, 0, 0, -1, -1, 0 };
void ser ()
{
hash[ 'R'-'A' ] = 'S';
hash[ 'S'-'A' ] = 'P';
hash[ 'P'-'A' ] = 'R';
int flag = 1;
for ( int i=0; i<n; i++ )
{
flag = 0;
for ( int j=0; j<r; j++ )
{
for ( int z=0; z<c; z++ )
{
int tr, tc;
for ( int x =0; x<4; x++ )
{
tr = j+move[x][0];
tc = z+move[x][1];
if ( tr>=0 && tr<r && tc>=0 && tc<c )
{
if ( hash[ map[tr][tc]-'A' ] == map[j][z] )
{
cmap[j][z] = map[tr][tc];
flag = 1;
break;
}
}
}
if ( x >= 4 )
{
cmap[j][z] = map[j][z];
}
}
}
cpy ();
if ( ! flag )
{
break;
}
}
}
int main ()
{
int t;
scanf ( "%d", &t );
while ( t -- )
{
scanf ( "%d%d%d", &r, &c, &n );
getchar ();
for ( int i=0; i<r; i++ )
{
for ( int j=0; j<c; j++ )
{
scanf ( "%c", &map[i][j] );
}
getchar ();
}
ser ();
for ( i=0; i<r; i++ )
{
for ( int j=0; j<c; j++ )
{
printf ( "%c", map[i][j] );
}
printf ( "\n" );
}
printf ( "\n" );
}
return 0;
}
摘要: 看题目是字符串处理,但实际上是欧拉回路
#include <stdio.h>#include <stdlib.h>#include <string.h>const int LEN = 30;struct HEAD{ int i... 阅读全文
摘要: 大数运算,用了模
#include<stdio.h>#include<stdlib.h>#include<string.h>const int OneNode = 1000000; //??????????????OneNodeconst int NodeLen = 6;... 阅读全文
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