POJ 2421 Constructing Roads
2015-05-08 23:33
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题意:给出邻接矩阵,求这个图的最小生成树
链接:http://poj.org/problem?id=2421
思路:最小生成树,因为给出的是邻接矩阵,求使任意两点之间最少花费
注意点:无
以下为AC代码:
链接:http://poj.org/problem?id=2421
思路:最小生成树,因为给出的是邻接矩阵,求使任意两点之间最少花费
注意点:无
以下为AC代码:
Run ID | User | Problem | Result | Memory | Time | Language | Code Length | Submit Time |
7260637 | 2421 | 1182 | Accepted | 1072K | 297MS | G++ | 1956B | 2010-07-26 21:01:17 |
#include <algorithm> #include <iostream> #include <climits> #include <iomanip> #include <cstdlib> #include <cstring> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <queue> #include <stack> #include <cmath> #include <ctime> #include <deque> #include <list> #include <map> #include <set> //#include <unordered_map> //#include <unordered_set> #define pb push_back #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 RS(s) scanf ( "%s", s ) #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 PL() printf ( "\n" ) #define PI(a) printf ( "%d", a ) #define PIL(a) printf ( "%d\n", a ) #define PSL(s) printf ( "%s\n", s ) #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 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) #pragma comment ( linker, "/STACK:1024000000,1024000000" ) 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) {} }; struct edge{ int to, wei; edge(){} edge( int _to, int _wei ) : to(_to), wei(_wei) {} }; bool cmp ( const node &a, const node &b ) { return a.cnt < b.cnt; } int ans; int m, n; int g[205][205]; node p[20005]; int fa[205]; int find ( int x ) { if ( x == fa[x] ){ return x; } else{ fa[x] = find ( fa[x] ); } } bool merge ( int u, int v ) { int x = find ( u ); int y = find ( v ); if ( x != y ){ fa[x] = y; return true; } else return false; } int main() { //int m, n; while ( RDI ( m ) != EOF ){ int k = 0; ans = 0; rep ( i, 0, 105 )fa[i] = i; REP ( i, 1, m ){ REP ( j, 1, m ){ RDI ( g[i][j] ); if ( j > i )p[k++] = node ( i, j, g[i][j] ); } } sort ( p, p + k, cmp ); RDI ( n ); int u, v; rep ( i, 0, n ){ RDII ( u, v ); merge ( u, v ); } rep ( i, 0, k ){ if ( merge ( p[i].x, p[i].y ) )ans += p[i].cnt; } PIL ( ans ); } return 0; }
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