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HDOJ 1233 还是畅通工程

2015-05-09 14:25 357 查看
题意:n个城镇,每个城镇之间要求相互联通,求至少需要多少长度路径

链接:http://acm.hdu.edu.cn/showproblem.php?pid=1233

思路:最小生成树模板

注意点:无

以下为AC代码:

Run IDSubmit TimeJudge StatusPro.IDExe.TimeExe.MemoryCode Len.LanguageAuthor
110219942014-07-14 10:45:06Accepted1233531MS348K1463 BC++luminous11
#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <string>
#include <vector>
#include <deque>
#include <list>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <cctype>
#include <climits>
#include <iomanip>
#include <cstdlib>
#include <algorithm>
//#include <unordered_map>
//#include <unordered_set>
#define ll long long
#define ull unsigned long long
#define all(x) (x).begin(), (x).end()
#define clr(a, v) memset( a , v , sizeof(a) )
#define pb push_back
#define RDI(a) scanf ( "%d", &a )
#define RDII(a, b) scanf ( "%d%d", &a, &b )
#define RDIII(a, b, c) scanf ( "%d%d%d", &a, &b, &c )
#define RS(s) scanf ( "%s", s )
#define PI(a) printf ( "%d", a )
#define PIL(a) printf ( "%d\n", a )
#define PII(a,b) printf ( "%d %d", a, b )
#define PIIL(a,b) printf ( "%d %d\n", a, b )
#define PIII(a,b,c) printf ( "%d %d %d", a, b, c )
#define PIIIL(a,b,c) printf ( "%d %d %d\n", a, b, c )
#define PL() printf ( "\n" )
#define PSL(s) printf ( "%s\n", s )
#define rep(i,m,n) for ( int i = m; i <  n; i ++ )
#define REP(i,m,n) for ( int i = m; i <= n; i ++ )
#define dep(i,m,n) for ( int i = m; i >  n; i -- )
#define DEP(i,m,n) for ( int i = m; i >= n; i -- )
#define repi(i,m,n,k) for ( int i = m; i <  n; i += k )
#define REPI(i,m,n,k) for ( int i = m; i <= n; i += k )
#define depi(i,m,n,k) for ( int i = m; i >  n; i += k )
#define DEPI(i,m,n,k) for ( int i = m; i >= n; i -= k )
#define READ(f) freopen(f, "r", stdin)
#define WRITE(f) freopen(f, "w", stdout)
using namespace std;
const double pi = acos(-1);
template <class T>
inline bool RD ( T &ret )
{
char c;
int sgn;
if ( c = getchar(), c ==EOF )return 0; //EOF
while ( c != '-' && ( c < '0' || c > '9' ) ) c = getchar();
sgn = ( c == '-' ) ? -1 : 1;
ret = ( c == '-' ) ? 0 : ( c - '0' );
while ( c = getchar() , c >= '0' && c <= '9' ) ret = ret * 10 + ( c - '0' );
ret *= sgn;
return 1;
}
inline void PD ( int x )
{
if ( x > 9 ) PD ( x / 10 );
putchar ( x % 10 + '0' );
}
const double eps = 1e-10;
const int dir[4][2] = { 1,0, -1,0, 0,1, 0,-1 };
struct node{
int x, y, cnt;
node(){}
node( int _x, int _y ) : x(_x), y(_y) {}
node( int _x, int _y, int _cnt ) : x(_x), y(_y), cnt(_cnt) {}
friend bool operator < ( const node &a, const node &b ){
return a.cnt < b.cnt;
}
};
int T;
int fa[105];
int ran[105];
vector<node>edge;
int find( int x )
{
return x == fa[x] ? x : find ( fa[x] );
}
inline int merge ( int i )
{
int x = find ( edge[i].x );
int y = find ( edge[i].y );
if ( x != y ){
if ( ran[x] < ran[y] ){
fa[x] = y;
ran[x] = max ( ran[x], ran[y] + 1 );
}
else{
fa[y] = x;
ran[y] = max ( ran[y], ran[x] + 1 );
}
return edge[i].cnt;
}
return 0;
}
void kruskal( int &ans )
{
int cnt = 1;
rep ( i, 0, edge.size() ){
int x = find ( edge[i].x );
int y = find ( edge[i].y );
if ( x != y ){
if ( ran[x] < ran[y] ){
fa[x] = y;
ran[x] = max ( ran[x], ran[y] + 1 );
}
else{
fa[y] = x;
ran[y] = max ( ran[y], ran[x] + 1 );
}
ans += edge[i].cnt;
cnt ++;
if ( cnt == T )return;
}
}
}
void init()
{
edge.clear();
rep ( i, 0, 105 )fa[i] = i;
clr ( ran, 0 );
}
int main()
{
int a, b, c;
while ( RDI ( T ) != EOF && T ){
init();
int k = T * ( T - 1 ) / 2;
rep ( i, 0, k ){
RDIII ( a, b, c );
edge.push_back( node ( a, b, c ) );
}
sort ( all( edge ) );
int ans = 0;
kruskal( ans );
PIL ( ans );
}
return 0;
}
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