题解 收藏家

传送门

是一个分层图网络流题
但是图全建出来会很大
发现每个时刻最多有四个点是有用的，所以其它点可以优化掉

点击查看代码

#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 1000010
#define ll long long
#define fir first
#define sec second
#define pb push_back
//#define int long long

char buf[1<<21], *p1=buf, *p2=buf;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf, 1, 1<<21, stdin)), p1==p2?EOF:*p1++)
inline int read() {
	int ans=0, f=1; char c=getchar();
	while (!isdigit(c)) {if (c=='-') f=-f; c=getchar();}
	while (isdigit(c)) {ans=(ans<<3)+(ans<<1)+(c^48); c=getchar();}
	return ans*f;
}

int n, m;
int a[N];

namespace task1{
	vector<int> uni[N];
	pair<int, int> link[N], rub[N];
	int head[N], size, cur[N], dep[N], id[3010][3010], s, t, ans, tot;
	struct edge{int to, next, val;}e[N<<1];
	inline void add(int s, int t, int w) {e[++size]={t, head[s], w}; head[s]=size;}
	inline int num(int x, int y) {return id[x][y];}
	bool bfs(int s, int t) {
		memset(dep, 0, sizeof(int)*(tot+10));
		queue<int> q;
		dep[s]=1; cur[s]=head[s];
		q.push(s);
		int u;
		while (q.size()) {
			u=q.front(); q.pop();
			for (int i=head[u],v; ~i; i=e[i].next) {
				v = e[i].to;
				if (e[i].val&&!dep[v]) {
					dep[v]=dep[u]+1;
					cur[v]=head[v];
					if (v==t) return 1;
					q.push(v);
				}
			}
		}
		return 0;
	}
	int dfs(int u, int in) {
		if (u==t||!in) return in;
		int rest=in, tem;
		for (int i=cur[u],v; ~i; cur[u]=i=e[i].next) {
			v = e[i].to;
			if (e[i].val&&dep[v]==dep[u]+1) {
				tem=dfs(v, min(rest, e[i].val));
				if (!tem) dep[v]=0;
				rest-=tem;
				e[i].val-=tem;
				e[i^1].val+=tem;
				if (!rest) break;
			}
		}
		return in-rest;
	}
	void solve() {
		// cout<<double(sizeof(head)*9)/1000/1000<<endl;
		size=1; ans=0;
		while (tot) id[rub[tot].fir][rub[tot].sec]=0, --tot;
		memset(head, -1, sizeof(head));
		s=++tot; t=++tot;
		for (int i=1; i<=n; ++i) uni[i].clear();
		for (int i=1,u,v; i<=m; ++i) {
			u=read(); v=read();
			link[i]={u, v};
			uni[u].pb(i); uni[v].pb(i);
		}
		for (int i=1; i<=n; ++i) {
			sort(uni[i].begin(), uni[i].end());
			uni[i].erase(unique(uni[i].begin(), uni[i].end()), uni[i].end());
			// cout<<"uni: "; for (auto it:uni[i]) cout<<it<<' '; cout<<endl;
			for (auto it:uni[i]) id[i][it]=++tot, rub[tot]={i, it};
			for (int j=1; j<uni[i].size(); ++j) add(num(i, uni[i][j-1]), num(i, uni[i][j]), a[i]), add(num(i, uni[i][j]), num(i, uni[i][j-1]), 0);
			if (uni[i].size()) add(s, num(i, uni[i][0]), 1), add(num(i, uni[i][0]), s, 0); //, cout<<"pos1"<<' '<<s<<' '<<uni[i][0]<<endl;
		}
		if (!uni[1].size()) {puts("1"); return ;}
		add(num(1, uni[1].back()), t, a[1]), add(t, num(1, uni[1].back()), 0);
		for (int i=1; i<=m; ++i) {
			add(num(link[i].fir, i), num(link[i].sec, i), 1), add(num(link[i].sec, i), num(link[i].fir, i), 0);
			add(num(link[i].sec, i), num(link[i].fir, i), 1), add(num(link[i].fir, i), num(link[i].sec, i), 0);
		}
		while (bfs(s, t)) ans+=dfs(s, INF); //, cout<<"link"<<endl;
		printf("%d\n", ans);
		// cout<<"size: "<<size<<endl;
		// cout<<ans<<endl;
	}
}

signed main()
{
	int T=read();
	while (T--) {
		n=read(); m=read();
		for (int i=1; i<=n; ++i) a[i]=read();
		task1::solve();
	}

	return 0;
}