P3183 [HAOI2016] food chain [topology / memory search]

Topic Source: Luo Gu

Title Description

Is a schematic view of a food web ecosystem, according to FIG Answer first small chain shown in FIG. Now you question and n and m species energy flow relationship, which find food. Species name is numbered from 1 to n relationship between the M-shaped flow of energy such as a1 b1a2 b2a3 b3 ...... am-1 bm-1am bm ai bi where ai represents the energy flow from the bi species species, a single note isolated not a biological food chain

Input and output formats

Input formats:

 

The first line of two integers n and m, each row next two rows ai bi integers m and m describe the relationship between the flow of energy. (Data of input data symbols to ensure biological characteristics, do not have duplicate energy flow relations) 1 <= N <= 100000 0 <= m <= 200000 Title answer guaranteed not burst int

 

Output formats:

 

An integer that is the number of food chains in the food web

 

Sample input and output

Input Sample # 1: 

10 16 
1 2 
1. 4 
1 10 
2. 3 
2. 5 
. 4. 3 
. 4. 5 
. 4. 8 
. 6. 5 
. 7. 6 
. 7. 9 
. 8. 5 
. 9. 8 
10. 6 
10. 7 
10. 9 
outputs Sample # 1: 

9 

Analysis: 
This problem may also dp, relatively water. 
I wrote a topological sort of naked, not much to say. 
Specifically look at the code. 
Note: d [] storage is through the food chain at a point number. Moreover, this question has eight test points are presented with isolated nodes, do not ask me how I know. 

Reference Code:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<string>
#include<cstdlib>
#include<queue>
#include<vector>
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define N 100010
#define MOD 2520
#define E 1e-12
using namespace std;
queue<int> q;
struct rec{
    int next,ver;
G} [N <<2 ];
 int head [N], TOT, n-, m, ING [N], CNT, D [N];
 BOOL V [N << 2 ];
 void the Add ( int X, int Y) 
{ 
    G [ ++ TOT] .ver = Y; 
    G [TOT] .next = head [X], head [X] = TOT; 
    ING [Y] ++; // can count on the degree of this time 
}
 int main () 
{ 
    Scanf ( " % D% D " , & n-, & m);
     for ( int I = . 1 ; I <= m; I ++ ) {
         int X, Y;
        Scanf ( " % D% D " , & X, & Y);
         IF (X == Y) Continue ; 
        the Add (X, Y); 
    } 
    for ( int I = . 1 ; I <= n-; I ++ )
         IF (ING [ I] == 0 ! && head [I] = 0 ) q.push (I), D [I] = . 1 ; // can head [] Analyzing an isolated point 
    the while (q.size ()) // bare topology 
    {
         int X = q.front (); q.pop ();
         IF (head [X] == 0 ) = CNT + D [X]; // still show a head batch of [] is determined that edge 
        for ( int I = head [X] ; i; i = G [I] .next) 
        { 
            int Y = G [I] .ver; 
            D [Y] + = D [X]; // accumulated through the food chain of the number of points 
            IF (Ing [Y] = = 0 ) q.push (Y); 
        } 
    } 
    COUT << CNT << endl;
     return  0 ; 
}