[HDU] 3549 Flow Problem [最大流][Dinic][读取优化]-优快云博客

本文介绍了一个使用Dinic算法解决网络流问题的方法，通过优化读取过程，显著提高了算法效率。针对加权有向图，该算法能够找到从源点到汇点的最大流值。

Problem Description
Network flow is a well-known difficult problem for ACMers. Given a graph, your task is to find out the maximum flow for the weighted directed graph.

Input
The first line of input contains an integer T, denoting the number of test cases.
For each test case, the first line contains two integers N and M, denoting the number of vertexes and edges in the graph. (2 <= N <= 15, 0 <= M <= 1000)
Next M lines, each line contains three integers X, Y and C, there is an edge from X to Y and the capacity of it is C. (1 <= X, Y <= N, 1 <= C <= 1000)

Output
For each test cases, you should output the maximum flow from source 1 to sink N.

Sample Input
2
3 2
1 2 1
2 3 1
3 3
1 2 1
2 3 1
1 3 1

Sample Output
Case 1: 1
Case 2: 2

link

题解

使用Dinic的复杂度是 O(E×V^2)

发现对其进行读取优化，效率立马提升3倍，可见读取优化对竞赛的重要性

这里写图片描述

#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
#include<algorithm>
#define INF 0x3f3f3f3f
#define MAX_V 20
using namespace std;
typedef long long LL;

const int MAXS = 5*1024*1024;
char buf[MAXS],bufout[MAXS],*ch,*chout;

void read(int &x){
    for(++ch;*ch<=32;++ch);
    for(x=0;*ch>='0';++ch) x=x*10+*ch-'0';
}

void out(int x){
    if(!x) *(++chout)='0';
    else{
        char *ch0=chout,*ch1=chout+1;
        while(x){
            *(++ch0)=x%10+'0';
            x/=10;
        }
        chout=ch0;
        while(ch1<=ch0) swap(*(ch1++),*(ch0--));
    }
    *(++chout)='\n';
}

void std_init(){
    ch=buf-1;
    chout=bufout-1;
    fread(buf,1,MAXS,stdin);
}

void std_out(){
    fwrite(bufout,1,chout-bufout+1,stdout);
}

/*---------------------------------------------------------------*/
struct edge{int to,cap,rev;};

vector<edge> G[MAX_V];
int lever[MAX_V];
int iter[MAX_V];

void add_edge(int x,int y,int cost){
    G[x].push_back((edge){y,cost,G[y].size()});
    G[y].push_back((edge){x,0,G[x].size()-1});
}
void del_graph(int N){for(int i=1;i<=N;i++) while(!G[i].empty()) G[i].pop_back();}

void bfs(int s){
    memset(lever,-1,sizeof(lever));
    queue<int> que;
    lever[s]=0;
    que.push(s);
    while(!que.empty()){
        int v=que.front();que.pop();
        for(int i=0;i<G[v].size();i++){
            edge &e=G[v][i];
            if(lever[e.to]<0&&e.cap>0){
                lever[e.to]=lever[v]+1;
                que.push(e.to);
            }
        }
    }
}

int dfs(int v,int t,int f){
    if(v==t) return f;
    for(int &i=iter[v];i<G[v].size();i++){
        edge &e=G[v][i];
        if(e.cap>0&&lever[v]<lever[e.to]){
            int d=dfs(e.to,t,min(f,e.cap));
            if(d>0){
                e.cap-=d;
                G[e.to][e.rev].cap+=d;
                return d;
            }
        }
    }
    return 0;
}

int max_flow(int s,int t){
    int flow=0,f;
    while(true){
        bfs(s);
        if(lever[t]<0) return flow;
        memset(iter,0,sizeof(iter));
        while(f=dfs(s,t,INF)) flow+=f;
    }
}

int main()
{
    std_init();
    int T,N,M;
    read(T);
    for(int t=1;t<=T;t++){
        read(N);read(M);
        int from,to,cost;
        for(int i=0;i<M;i++){
            read(from);read(to);read(cost);
            add_edge(from,to,cost);
        }
        printf("Case %d: %d\n",t,max_flow(1,N));
        del_graph(N);
    }
    return 0;
}