常州 d??

回来了！

今天（？发出来的时候可能已经是第二天了吧

真的爆零了qaq

挺难的 最高分只有100 而且t2t3t4一个人都没拿分qaq

这段时间在写网络流 唉博客还是不要写太水了 抄一点网络流代码上来

EK不写了 dinic会就行了

板子题洛谷p3376

dinic做法

#include<bits/stdc++.h>
#define int long long
using namespace std;
const int N = 210, M = 1e4+10;
int n, m, s, t, tot = 1, ver[M], h[N], nxt[M], edg[M], d[N], now[N], fl, inf = 0x7fffffff, maxf;
queue < int > q, clr;
void add(int u, int v, int w){
    ver[++tot] = v; edg[tot] = w, nxt[tot] = h[u];
    h[u] = tot;
    ver[++tot] = u; edg[tot] = 0, nxt[tot] = h[v];
    h[v] = tot;
}
bool bfs(){
    memset(d, 0, sizeof d); q = clr;
    q.push(s); d[s] = 1, now[s] = h[s];
    while (q.size()){
        int x = q.front(); q.pop();
        for (int i = h[x]; i; i = nxt[i]){
            if (!edg[i] || d[ver[i]]) continue;
            q.push(ver[i]);
            now[ver[i]] = h[ver[i]];
            d[ver[i]] = d[x] + 1;
            if (ver[i] == t) return true;
        }
    }
    return false;
}
int dinic(int x, int flw){
    if (x == t) return flw;
    int rst = flw;
    for (int i = now[x]; i; i = nxt[i]){
        now[x] = i;
        if (!edg[i] || d[ver[i]] != d[x] + 1) continue;
        int k = dinic(ver[i], min(rst, edg[i]));
        if (!k) d[ver[i]] = 0;
        edg[i] -= k; edg[i ^ 1] += k;
        rst -= k;
    }
    return flw - rst;
}
signed main(){
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    cin >> n >> m >> s >> t;
    for (int i = 1; i <= m; ++i){
        int u, v, w; cin >> u >> v >> w;
        add(u, v, w);
    }
    while (bfs()) while (fl = dinic(s, inf)) maxf += fl;
    cout << maxf << "\n";
    return 0;
}

dinic优化ISAP做法

#include<bits/stdc++.h>
#define int long long
using namespace std;
const int N = 210, M = 1e4+10;
int n, m, s, t, tot = 1, ver[M], h[N], nxt[M], edg[M], dep[N], gap[N], maxf, inf = 0x7fffffff, cur[N];
queue < int > q;
void add(int u, int v, int w){
    ver[++tot] = v; nxt[tot] = h[u], edg[tot] = w;
    h[u] = tot;
    ver[++tot] = u; nxt[tot] = h[v], edg[tot] = 0;
    h[v] = tot; 
}
void bfs(){
    memset(dep, -1, sizeof(dep));
    q.push(t); gap[dep[t] = 0] = 1;
    while (!q.empty()){
        int x = q.front(); q.pop();
        for (int i = h[x]; i; i = nxt[i]){
            if (dep[ver[i]] != -1) continue;
            q.push(ver[i]); dep[ver[i]] = dep[x] + 1;
            ++gap[dep[ver[i]]];
        }
    }
}
int dfs(int x, int flw){
    if (x == t) return flw;
    int rst = flw;
    for (int i = cur[x]; i; i = nxt[i]){
        cur[x] = i;
        if (!edg[i] || dep[ver[i]] != dep[x] - 1) continue;
        int k = dfs(ver[i], min(rst, edg[i]));
        edg[i] -= k, edg[i ^ 1] += k;
        rst -= k;
        if (!rst) return flw;
    } 
    --gap[dep[x]];
    if (!gap[dep[x]]) dep[s] = n + 1;
    ++gap[++dep[x]];
    return flw - rst;
}
int isap(){
    bfs();
    while (dep[s] < n){
        memcpy(cur, h, sizeof(cur));
        maxf += dfs(s, inf);
    }
    return maxf;
}
signed main(){
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    cin >> n >> m >> s >> t;
    for (int i = 1; i <= m; ++i){
        int u, v, w; cin >> u >> v >> w;
        add(u, v, w);
    }
    cout << isap() << "\n";
    return 0;
}

费用流板子题洛谷p3381

dinic+spfa

#include<bits/stdc++.h>
//#define int long long
using namespace std;
const int N = 5010, M = 200010, inf = 0x3f3f3f3f;
int m, n, k, tot = 1, h[N], ver[M], nxt[M], edg[M], cst[M], dis[N], cur[N], s, t, ans, maxf, fl;
bool vis[N], v[N];
queue < int > q, clr;
void add(int u, int v, int w, int c){
	ver[++tot] = v; nxt[tot] = h[u], edg[tot] = w, cst[tot] = c;
	h[u] = tot;
	ver[++tot] = u; nxt[tot] = h[v], edg[tot] = 0, cst[tot] = -c;
	h[v] = tot;
}
bool spfa(){
	q = clr;
	memset(dis, 0x3f, sizeof(dis));
	memset(vis, 0, sizeof vis);
	q.push(s); dis[s] = 0, vis[s] = 1;
	while (q.size()){
		int x = q.front(); q.pop(); vis[x] = 0;
		for (int i = h[x]; i; i = nxt[i]){
			if (!edg[i] || dis[ver[i]] <= dis[x] + cst[i]) continue;
			dis[ver[i]] = dis[x] + cst[i];
			if (!vis[ver[i]]) vis[ver[i]] = 1, q.push(ver[i]);
		}
	}
	return dis[t] != inf;
}
int dinic(int x, int flw){
	v[x] = 1;
	if (x == t) return flw;
	int rst = flw;
	for (int i = cur[x]; i; i = nxt[i]){
		cur[x] = i;
		if (!edg[i] || dis[ver[i]] != dis[x] + cst[i] || v[ver[i]]) continue; 
		v[ver[i]] = 1;
		int k = dinic(ver[i], min(rst, edg[i]));
		edg[i] -= k, edg[i ^ 1] += k;
		rst -= k; ans += k * cst[i];
	}
	return flw - rst;
}
signed main() {
    //ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    cin >> n >> m >> s >> t;
    for (int i = 1; i <= m; ++i){
    	int u, v, w, c; cin >> u >> v >> w >> c;
    	add(u, v, w, c);
	}
	while (spfa()){
		memcpy(cur, h, sizeof(h));
		memset(v, 0, sizeof(v));
		while ((fl = dinic(s, inf)) > 0) maxf += fl;
	}
	cout << maxf << " " << ans << "\n";
    return 0;
}

其他一些例题 不放代码了

洛谷p4452 

洛谷p2053

洛谷p2045

如果有可能说不定到时候会写网络流学习笔记...?（大概率是不可能的