• 题意
    • 最左边的点到最右边的点的最大流
    • n,m<=1e5
    • 双向边
  • 题解
      • 直接跑dinic会t
      • 需要弧优化
      • 然后双向边不需要建立反向弧,直接正反建边即可
  • 代码
    • #include<algorithm>
      #include <iostream>
      #include   <fstream>
      #include  <cstdlib>
      #include  <cstring>
      #include  <cassert>
      #include   <cstdio>
      #include   <vector>
      #include   <string>
      #include    <cmath>
      #include    <queue>
      #include    <stack>
      #include      <set>
      #include      <map>
      using namespace std;
      #define P(a,b,c) make_pair(a,make_pair(b,c))
      #define rep(i,a,n) for (int i=a;i<=n;i++)
      #define per(i,a,n) for (int i=n;i>=a;i--)
      #define CLR(vis) memset(vis,0,sizeof(vis))
      #define MST(vis,pos) memset(vis,pos,sizeof(vis))
      #define pb push_back
      #define mp make_pair
      #define all(x) (x).begin(),(x).end()
      #define fi first
      #define se second
      #define SZ(x) ((int)(x).size())
      typedef pair<int,pair<int,int> >pii;
      typedef long long ll;
      typedef unsigned long long ull;
      const ll mod = 1000000007;
      const int INF = 0x3f3f3f3f;
      ll gcd(ll a, ll b) {
      	return b ? gcd(b, a%b) : a;
      }
      template<class T>inline void gmax(T &A, T B) {
      	(A<B)&&(A=B);//if(B>A)A=B;
      }
      template<class T>inline void gmin(T &A, T B) {
      	(A>B)&&(A=B);//if(B<A)A=B;
      }
      template <class T>
      inline bool scan_d(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 outt(int x) {
      	if(x>9) outt(x/10);
      	putchar(x%10+'0');
      }
      
      #define inf 0x3f3f3f3f
      const int maxn = 100010;
      const int maxm = 1000100;
      struct fuck {
      	int v, w, ne;
      }ed[maxm];
      int n, m, cnt;
      int head[maxn], dis[maxn], cur[maxn];
      void init() {
      	cnt = 0;
      	memset(head, -1, sizeof(head));
      }
      void add(int u, int v, int w) {
      	ed[cnt].v = v; ed[cnt].w = w;
      	ed[cnt].ne = head[u]; head[u] = cnt++;
      //	ed[cnt].v = u, ed[cnt].w = 0;
      //	ed[cnt].ne = head[v]; head[v] = cnt++;
      }
      int bfs(int st, int en) {
      	queue<int>q;
      	memset(dis, 0, sizeof(dis));
      	dis[st] = 1;
      	q.push(st);
      	while (!q.empty()) {
      		int u = q.front(); q.pop();
      		if (u == en)return 1;
      		for (int s = head[u]; ~s; s = ed[s].ne) {
      			int v = ed[s].v;
      			if (dis[v] == 0 && ed[s].w > 0) {
      				dis[v] = dis[u] + 1; q.push(v);
      			}
      		}
      	}
      	return dis[en] != 0;
      }
      int dfs(int st, int en, int flow) {
      	int ret = flow, a;
      	if (st == en || flow == 0)return flow;
      	for (int &s = cur[st]; ~s; s = ed[s].ne) {
      		int v = ed[s].v;
      		if (dis[v] == dis[st] + 1 && (a = dfs(v, en, min(ret, ed[s].w)))) {
      			ed[s].w -= a;
      			ed[s ^ 1].w += a;
      			ret -= a;
      			if (!ret)break;
      		}
      	}
      	if (ret == flow)dis[st] = 0;
      	return flow - ret;
      }
      int dinic(int st, int en) {
      	int ans = 0;
      	while (bfs(st, en)) {
      		for (int s = 1; s <= n; s++) 
      			cur[s] = head[s];
      		ans += dfs(st, en, inf);
      	}
      	return ans;
      }
      int main() {
      	int T;
      	scan_d(T);
      	int s,t;
      	while(T--) {
      		init();
      		scanf("%d%d", &n, &m);
      		int x;
      		int maxx = 0;
      		int minx = INF;
      		rep(i,1,n) {
      			scan_d(x);
      			if (x<minx) {
      				minx=x;
      				s=i;
      			}
      			if (x>maxx) {
      				maxx=x;
      				t=i;
      			}
      			scan_d(x);
      		}
      		int u,v,w;
      		rep(i,1,m){
      			scanf("%d%d%d", &u,&v,&w);
      			add(u,v,w);
      			add(v,u,w);
      		}
      		int maxflow = dinic(s,t);
      		printf("%d\n", maxflow);
      	}
      	return 0;
      }

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