POJ 3164 最小樹形圖

這題才是最小樹形圖的基礎題，題意就不贅述了，敲這道題的時候發現一個很坑的情況。

因為平時輸入的時候用慣了輸入優化，所以對於這些輸入我一般都直接上輸入優化的，但是這道題讓我T了20次之後我才發現輸入優化居然是T的原因，我改成scanf後就A掉了。

比如下面那段代碼的注釋處，改成輸入優化就T了。

不解，求解答。

#include <set>   
#include <map>   
#include <stack>   
#include <cmath>   
#include <queue>   
#include <cstdio>   
#include <string>   
#include <vector>   
#include <iomanip>   
#include <cstring>   
#include <iostream>   
#include <algorithm>   
#define Max 2505   
#define FI first   
#define SE second   
#define ll long long   
#define PI acos(-1.0)   
#define inf 0x7fffffff   
#define LL(x) ( x << 1 )   
#define bug puts("here")   
#define PII pair<int,int>   
#define RR(x) ( x << 1 | 1 )   
#define mp(a,b) make_pair(a,b)   
#define mem(a,b) memset(a,b,sizeof(a))   
#define REP(i,s,t) for( int i = ( s ) ; i <= ( t ) ; ++ i )   
  
using namespace std;  
  
inline void RD(int &ret) {  
    char c;  
    int flag = 1 ;  
    do {  
        c = getchar();  
        if(c == '-')flag = -1 ;  
    } while(c < '0' || c > '9') ;  
    ret = c - '0';  
    while((c=getchar()) >= '0' && c <= '9')  
        ret = ret * 10 + ( c - '0' );  
    ret *= flag ;  
}  
inline void OT(int a) {  
    if(a >= 10)OT(a / 10) ;  
    putchar(a % 10 + '0') ;  
}  
  
inline void RD(double &ret) {  
    char c ;  
    int flag = 1 ;  
    do {  
        c = getchar() ;  
        if(c == '-')flag = -1 ;  
    } while(c < '0' || c > '9') ;  
    ll n1 = c - '0' ;  
    while((c = getchar()) >= '0' && c <= '9') {  
        n1 = n1 * 10 + c - '0' ;  
    }  
    ll n2 = 1 ;  
    while((c = getchar()) >= '0' && c <= '9') {  
        n1 = n1 * 10 + c - '0' ;  
        n2 *= 10 ;  
    }  
    ret = flag * (double)n1 / (double)(n2) ;  
}  
/*********************************************/  
  
#define N 1005   
struct PP{  
    double x , y ;  
}p[N] ;  
struct kdq{  
    int s , e ;  
    double l ;  
}ed[N * N] ;  
int n , m ;  
double getdis(int i ,int j){  
    return sqrt((p[i].x - p[j].x) * (p[i].x - p[j].x) + (p[i].y - p[j].y) * (p[i].y - p[j].y)) ;  
}  
int pre[N] , vis[N] , id[N] ;  
double in[N] ;  
double Directed_MST(int root ,int V ,int E){  
    double ret = 0 ;  
    while(1){  
        for (int i = 1 ; i < V ; i ++ )in[i] = inf ;  
        for (int i = 0 ; i < E ; i ++ ){  
            int s = ed[i].s ;  
            int e = ed[i].e ;  
            if(in[e] > ed[i].l && s != e){  
                pre[e] = s ;  
                in[e] = ed[i].l ;  
            }  
        }  
        for (int i = 1 ; i < V ; i ++ ){  
            if(i == root)continue ;  
            if(in[i] == inf)return -1 ;  
        }  
        int cntnode = 1 ;  
        mem(vis , -1) ;  
        mem(id ,-1) ;  
        in[root] = 0 ;  
        for (int i = 1 ; i < V ; i ++ ){  
            ret += in[i] ;  
            int v = i ;  
            while(vis[v] != i && id[v] == -1 && v != root){  
                vis[v] = i ;  
                v = pre[v] ;  
            }  
            if(v != root && id[v] == -1){  
                for (int u = pre[v] ; u != v ; u = pre[u]){  
                    id[u] = cntnode ;  
                }  
                id[v] = cntnode ++ ;  
            }  
        }  
        if(cntnode == 1)break ;  
        for (int i = 1 ; i < V ; i ++ ){  
            if(id[i] == -1)id[i] = cntnode ++ ;  
        }  
        for (int i = 0 ; i < E ; i ++ ){  
            int s = ed[i].s ;  
            int e = ed[i].e ;  
            ed[i].s = id[s] ;  
            ed[i].e = id[e] ;  
            if(id[s] != id[e])ed[i].l -= in[e] ;  
        }  
        V = cntnode ;  
        root = id[root] ;  
    }  
    return ret ;  
}  
  
int main() {  
    while(scanf("%d%d",&n,&m) != EOF){  
        for (int i = 1 ; i <= n ;i ++ ){  
            scanf("%lf%lf",&p[i].x ,&p[i].y) ;  
        }  
        for (int i = 0 ; i < m ; i ++ ){  
//            RD(ed[i].s) ; RD(ed[i].s) ;   
            scanf("%d%d",&ed[i].s ,&ed[i].e) ;  
            if(ed[i].s != ed[i].e)//消除自環   
            ed[i].l = getdis(ed[i].s , ed[i].e) ;  
            else  
            ed[i].l = inf ;  
        }  
        double ans = Directed_MST(1 , n + 1 , m) ;  
        if(ans == -1)printf("poor snoopy\n") ;  
        else  
        printf("%.2f\n",ans) ;  
    }  
    return 0 ;  
}