POJ 1251 / HDOJ 1301 Jungle Roads

题意：已知一张图，求出使图任意两点之间有路径的最小花费

链接：http://poj.org/problem?id=1251         http://acm.hdu.edu.cn/showproblem.php?pid=1301

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.LanguageAuthor110010532014-07-11 14:57:58Accepted13010MS292K1378 BC++luminous11

Run IDUserProblemResultMemoryTimeLanguageCode LengthSubmit Time14173750luminous111251Accepted724K0MSG++3666B2015-05-08 08:01:07

#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[105];int ran[105];int n;void init(){    g.clear();    clr ( ran, 0 );    rep ( i, 0, 105 )fa[i] = i;}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(){    ios::sync_with_stdio( false );    while ( cin >> n && n ){        char be, en;        int k;        int d;        init();        while ( -- n ){            cin >> be >> k;            rep ( i, 0, k ){                cin >> en >> d;                g.push_back( node( be-'A', en-'A', d ) );            }        }        int ans = 0;        kruskal( ans );        PIL ( ans );    }    return 0;}