POJ 3164 最小树形图
这题才是最小树形图的基础题，题意就不赘述了，敲这道题的时候发现一个很坑的情况。
因为平时输入的时候用惯了输入优化，所以对于这些输入我一般都直接上输入优化的，但是这道题让我T了20次之后我才发现输入优化居然是T的原因，我改成scanf后就A掉了。
比如下面那段代码的注释处，改成输入优化就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)//消除自环
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 ;
}