POJ 1287 Networking
来源:互联网 发布:linux cst破解 编辑:程序博客网 时间:2024/05/23 11:15
题意:已知在一张无向图中的每条边的权重,求将图中任意两点联通的最小花费
链接:http://poj.org/problem?id=1287
思路:最小生成树模板题,数据范围小prim/kruskal均可
注意点:无
以下为AC代码:
Run IDUserProblemResultMemoryTimeLanguageCode LengthSubmit Time14175050luminous111287Accepted712K16MSG++3555B2015-05-08 17:22:56#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; }};vector<node>g;int fa[1005];int ran[1005];void init(){ rep ( i, 0, 1005 )fa[i] = i; clr ( ran, 0 ); g.clear();}int find ( int x ){ return x == fa[x] ? x : find ( fa[x] );}void kruskal( int &ans ){ sort ( all( g ) ); rep ( i, 0, g.size() ){ int x = find ( g[i].x ); int y = find ( g[i].y ); if ( x != y ){ if ( ran[x] < ran[y] ){ fa[x] = y; ran[y] = max ( ran[y], ran[x] + 1 ); } else{ fa[y] = x; ran[x] = max ( ran[x], ran[y] + 1 ); } ans += g[i].cnt; } }}int main(){ int n, m; while ( RDII ( n, m ) != EOF ){ if ( n == 0 )break; init(); int a, b, c; rep ( i, 0, m ){ RDIII ( a, b, c ); g.push_back( node( a, b, c ) ); } int ans = 0; kruskal( ans ); PIL ( ans ); } return 0;}
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