HDU 4280:http://acm.hdu.edu.cn/showproblem.php?pid=4280

題意：

　　比較裸的最大流題目，就是這是個無向圖，並且比較卡時間。

思路：

　　是這樣的，由於是無向圖，所以addedge 的反邊容量直接設為原始流量。然後還可以優化搜索的方向，bfs可以從t到s跑，dfs可以從s到t跑，這樣快。

//#pragma GCC optimize(3)
//#pragma comment(linker, "/STACK:102400000,102400000")  //c++
// #pragma GCC diagnostic error "-std=c++11"
// #pragma comment(linker, "/stack:200000000")
 
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
// #pragma GCC optimize("-fdelete-null-pointer-checks,inline-functions-called-once,-funsafe-loop-optimizations,-fexpensive-optimizations,-foptimize-sibling-calls,-ftree-switch-conversion,-finline-small-functions,inline-small-functions,-frerun-cse-after-loop,-fhoist-adjacent-loads,-findirect-inlining,-freorder-functions,no-stack-protector,-fpartial-inlining,-fsched-interblock,-fcse-follow-jumps,-fcse-skip-blocks,-falign-functions,-fstrict-overflow,-fstrict-aliasing,-fschedule-insns2,-ftree-tail-merge,inline-functions,-fschedule-insns,-freorder-blocks,-fwhole-program,-funroll-loops,-fthread-jumps,-fcrossjumping,-fcaller-saves,-fdevirtualize,-falign-labels,-falign-loops,-falign-jumps,unroll-loops,-fsched-spec,-ffast-math,Ofast,inline,-fgcse,-fgcse-lm,-fipa-sra,-ftree-pre,-ftree-vrp,-fpeephole2",3)
 


#include <algorithm>
#include  <iterator>
#include  <iostream>
#include   <cstring>
#include   <cstdlib>
#include   <iomanip>
#include    <bitset>
#include    <cctype>
#include    <cstdio>
#include    <string>
#include    <vector>
#include     
 
<stack>
#include     <cmath>
#include     <queue>
#include      <list>
#include       <map>
#include       <set>
#include   <cassert>

using namespace std;
#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue



typedef long long ll;
typedef unsigned long long ull;
//typedef __int128 bll;
typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;
typedef pair<int,pii> p3;

//priority_queue<int> q;//這是一個大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//這是一個小根堆q
#define fi first
#define se second
//#define endl ‘\n‘

#define OKC ios::sync_with_stdio(false);cin.tie(0)
#define FT(A,B,C) for(int A=B;A <= C;++A)  //用來壓行
#define REP(i , j , k)  for(int i = j ; i <  k ; ++i)
#define max3(a,b,c) max(max(a,b), c);
//priority_queue<int ,vector<int>, greater<int> >que;

const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000;  //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const int mod = 1e9+7;
const double esp = 1e-8;
const double PI=acos(-1.0);
const double PHI=0.61803399;    //黃金分割點
const double tPHI=0.38196601;


template<typename T>
inline T read(T&x){
    x=0;int f=0;char ch=getchar();
    while (ch<‘0‘||ch>‘9‘) f|=(ch==‘-‘),ch=getchar();
    while (ch>=‘0‘&&ch<=‘9‘) x=x*10+ch-‘0‘,ch=getchar();
    return x=f?-x:x;
}


/*-----------------------showtime----------------------*/

            struct edge{
                int u,v,cap;
                edge(){}
                edge(int u,int v,int cap):
                   u(u),v(v),cap(cap) {}
            }es[200009];

            int tot,s,t;
            vector<int>tab[100009];
            int dis[100009],cur[100009];
            void addedge(int u,int v,int cap){
                tab[u].pb(tot);
                es[tot++] = edge(u,v,cap);
                tab[v].pb(tot);
                es[tot++] = edge(v,u,cap);
            }

            bool bfs(){
                queue<int>q;q.push(t);
                memset(dis,inf,sizeof(dis));
                dis[t] = 1;
                while(!q.empty()){
                    int h = q.front();q.pop();
                    for(int i=0; i<tab[h].size(); i++){
                        edge & e = es[tab[h][i]];

                        if(es[tab[h][i]^1].cap>0 && dis[e.v] >= inf){
                            dis[e.v] = dis[h] + 1;
                            q.push(e.v);
                        }
                    }
                }
                return dis[s] < inf;
            }
            int dfs(int x,int maxflow){
                if(x == t || maxflow == 0) return maxflow;
                for(int i=cur[x]; i<tab[x].size(); i++){
                    cur[x] = i;
                    edge & e = es[tab[x][i]];
                    if(dis[e.v] == dis[x] - 1 && e.cap > 0){
                        int flow = dfs(e.v, min(maxflow, e.cap));
                        if(flow){
                            e.cap -= flow;
                            es[tab[x][i] ^ 1].cap += flow;
                            return flow;
                        }
                    }
                }
                return 0;
            }
            int dinic(){
                int ans = 0;
                while(bfs()){
                    int flow;
                    memset(cur,0,sizeof(cur));
                    do{
                        flow = dfs(s,inf);
                        if(flow) ans += flow;
                    }while(flow);
                }
                return ans;
            }
int main(){
            int T;scanf("%d", &T);
            while(T--){
                int n,m,x,y,nx = inf,ny = -inf;    
                // scanf("%d%d",&n, &m);
                read(n),read(m);
                for(int i=1; i<=n; i++)tab[i].clear();
                tot = 0;
                for(int i=1; i<=n; i++){
                    int x,y;
                    // scanf("%d%d", &x, &y);
                    read(x),read(y);
                    if(x < nx) s = i, nx = x;
                    if(x > ny) t = i, ny = x;
                }
                // debug(s);
                //debug(t);
                for(int i=1; i<=m; i++){
                    int u,v,w;
                    // scanf("%d%d%d", &u, &v, &w);
                    read(u),read(v),read(w);
                    addedge(u, v, w);
                   // addedge(v, u, w);
                }
                printf("%d\n", dinic());
            }

            return 0;
}

HDU 4280 Island Transport（無向圖最大流）