HDOJ 1532 Drainage Ditches

题意：有一个n个点和m条边的网络，每条边都有流量上限，求从点1到点n的最大流量

链接：http://acm.hdu.edu.cn/showproblem.php?pid=1532

思路：网络流dinic模板

注意点：无

以下为AC代码：

Run IDSubmit TimeJudge StatusPro.IDExe.TimeExe.MemoryCode Len.LanguageAuthor131511342015-03-16 22:56:54Accepted15320MS1684K4234 BC++luminous11

#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 edge{    int to, f, e;    edge(){}    edge( int _to, int _f, int _e ) : to(_to), f(_f), e(_e) {}};const int MAXN = 205;const int INF = 0x3f3f3f3f;int n, m;vector<edge> g[MAXN];bool vis[MAXN];int dis[MAXN];queue<int> q;int be, en;void init(){    while ( ! q.empty() ) q.pop();    rep ( i, 0, MAXN ) g[i].clear();}int dfs ( int u, int f ){    if ( u == en ){        return f;    }    else{        int ret = 0;        rep ( i, 0, g[u].size() ){            if ( f == 0 )break;            if ( g[u][i].f != g[u][i].e && dis[g[u][i].to] == dis[u] + 1 ){                int dd = dfs ( g[u][i].to, min ( g[u][i].e - g[u][i].f, f ) );                g[u][i].f += dd;                f -= dd;                ret += dd;            }        }        return ret;    }}void bfs (){    vis[be] = 1;    q.push ( be );    while ( ! q.empty() ){        int u = q.front();        q.pop();        rep ( i, 0, g[u].size() ){            if ( ! vis[g[u][i].to] && g[u][i].e != g[u][i].f ){                dis[g[u][i].to] = dis[u] + 1;                vis[g[u][i].to] = 1;                q.push ( g[u][i].to );            }        }    }}void print(){    rep ( i, 0, m ){        rep ( j, 0, g[i].size() ){            cout << " i : " << i << " to : " << g[i][j].to << " f : " << g[i][j].f << " e : " << g[i][j].e << endl;        }    }}int maxflow(){    int f = 0;    while ( 1 ){        clr ( vis, 0 );        clr ( dis, 0 );        bfs();        if ( vis[en] == false )            return f;        f += dfs ( be, INF );    }}int main(){    int a, b, c;    while ( RDII ( n, m ) != EOF ){        init();        be = 1;        en = m;        rep ( i, 0, n ){            RDIII ( a, b, c );            g[a].push_back ( edge ( b, 0, c ) );        }        int ans = maxflow();        PIL ( ans );    }    return 0;}