POJ 1258 Agri-Net

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题意：已知有n个村庄和每个村庄之间的距离，求将所有村庄全部连接起来的最短距离

链接：http://poj.org/problem?id=1258

思路：最小生成树模板

注意点：无

以下为AC代码：

Run IDUserProblemResultMemoryTimeLanguageCode LengthSubmit Time14177078luminous111258Accepted704K0MSG++3438B2015-05-09 12:10:52

#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) {}};int g[105][105];int prim ( int n ){    int ans = 0;    int dis[105];    int vis[105];    clr ( vis, 0 );    clr ( dis, 0 );    REP ( i, 1, n ){        dis[i] = g[1][i];    }    rep ( k, 0, n ){        int _min = 0x3f3f3f3f;        int p = -1;        REP ( i, 1, n ){            if ( ! vis[i] && _min > dis[i] ){                _min = dis[i];                p = i;            }        }        if ( p == -1 )return -1;        ans += dis[p];        vis[p] = 1;        REP ( i, 1, n ){            if ( ! vis[i] && dis[i] > g[p][i] ){                dis[i] = g[p][i];            }        }    }    return ans;}int main(){    int n;    while ( RDI ( n ) != EOF ){        REP ( i, 1, n ){            REP ( j, 1, n ){                RDI ( g[i][j] );            }        }        int mst = prim( n );        PIL ( mst );    }    return 0;}

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