HDU 1532 Drainage Ditches

题目链接：https://vjudge.net/problem/HDU-1532

题目大意

　　给定 m 个点，n 条边，以及每条边的容量 c，求从原点 1 到汇点 m 的最大流。

分析：

　　网络流模板题。

代码如下

#include <bits/stdc++.h>
using namespace std;
 
#define INIT() std::ios::sync_with_stdio(false);std::cin.tie(0);
#define Rep(i,n) for (int i = 0; i < (n); ++i)
#define For(i,s,t) for (int i = (s); i <= (t); ++i)
#define rFor(i,t,s) for (int i = (t); i >= (s); --i)
#define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)
#define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
#define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
#define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)
 
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
 
#define LOWBIT(x) ((x)&(-x))
 
#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())
 
#define ms0(a) memset(a,0,sizeof(a))
#define msI(a) memset(a,inf,sizeof(a))
#define msM(a) memset(a,-1,sizeof(a))
#define mcy(d,s) memcpy(d, s, sizeof(s))

#define MP make_pair
#define PB push_back
#define ft first
#define sd second
 
template<typename T1, typename T2>
istream &operator>>(istream &in, pair<T1, T2> &p) {
    in >> p.first >> p.second;
    return in;
}
 
template<typename T>
istream &operator>>(istream &in, vector<T> &v) {
    for (auto &x: v)
        in >> x;
    return in;
}
 
template<typename T1, typename T2>
ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
    out << "[" << p.first << ", " << p.second << "]" << "
";
    return out;
}

inline int gc(){
    static const int BUF = 1e7;
    static char buf[BUF], *bg = buf + BUF, *ed = bg;
    
    if(bg == ed) fread(bg = buf, 1, BUF, stdin);
    return *bg++;
} 

inline int ri(){
    int x = 0, f = 1, c = gc();
    for(; c<48||c>57; f = c=='-'?-1:f, c=gc());
    for(; c>47&&c<58; x = x*10 + c - 48, c=gc());
    return x*f;
}
 
typedef long long LL;
typedef unsigned long long uLL;
typedef pair< double, double > PDD;
typedef pair< int, int > PII;
typedef pair< string, int > PSI;
typedef set< int > SI;
typedef vector< int > VI;
typedef map< int, int > MII;
typedef pair< LL, LL > PLL;
typedef vector< LL > VL;
typedef vector< VL > VVL;
const double EPS = 1e-10;
const LL inf = 0x7fffffff;
const LL infLL = 0x7fffffffffffffffLL;
const LL mod = 1e9 + 7;
const int maxN = 2e2 + 7;
const LL ONE = 1;
const LL evenBits = 0xaaaaaaaaaaaaaaaa;
const LL oddBits = 0x5555555555555555;

struct Vertex{
    VI edges;
};

struct Edge{
    int from, to;
    int r; // 残余容量
};

Edge e[maxN]; 
Vertex v[maxN];
int elen;
int n, m;

int pre[maxN]; //  // 从 s - t 中的一个可行流中, Pre[i]表示指向节点 i 的边
bool vis[maxN]; // 标记一个点是否被访问过

// bfs寻找增广，如果找到就返回true，没找到就返回false
bool bfs(int S, int T) {
    queue< int > Q;
    msM(pre);
    ms0(vis);
    vis[S] = 1;
    Q.push(S);
    
    while(!Q.empty()) {
        int tmpQ = Q.front();
        Q.pop();
        
        foreach(i, v[tmpQ].edges) {
            int to = e[*i].to;
            if(e[*i].r == 0 || vis[to]) continue;
            vis[to] = true;
            pre[to] = *i;
            if(to == T) return true;
            Q.push(to);
        }
    }
    return false;
}

// EdmondsKarp算法求最大流
// S 源点 
// T 汇点 
int maxFlow(int S, int T) {
    int ret = 0;
    
    // 不断寻找增广路 
    while(bfs(S, T)) {
        // 找可行流中残余流量最小的边 
        int t = T;
        int mi = inf;
        while(pre[t] != -1) {
            mi = min(mi, e[pre[t]].r);
            t = e[pre[t]].from;
        }
        
        // 更新可行流的所有边，减去最小残余流量 
        ret += mi;
        t = T;
        while(pre[t] != -1) {
            e[pre[t]].r -= mi;
            t = e[pre[t]].from;
        }
    }
    return ret;
}

int main(){
    INIT();
    while(cin >> n >> m) {
        elen = 0;
        For(i, 1, m) v[i].edges.clear();
        
        Rep(i, n) {
            int S, E, C;
            cin >> S >> E >> C;
            e[elen].from = S;
            e[elen].to = E;
            e[elen].r = C;
            v[S].edges.PB(elen++);
        }
        
        cout << maxFlow(1, m) << endl;
    }
    return 0;
}