BZOJ4873 [Shoi2017]寿司餐厅 【最大权闭合子图】

题目链接

BZOJ4873

题解

题意很鬼畜，就可以考虑网络流【雾】
然后就会发现这是一个裸的最大权闭合子图

就是注意要离散化一下代号

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 11005,maxm = 8000005,INF = 1000000000;
inline int read(){
	int out = 0,flag = 1; char c = getchar();
	while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
	while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
	return out * flag;
}
int h[maxn],ne = 1;
struct EDGE{int to,nxt,f;}ed[maxm];
inline void build(int u,int v,int f){
	ed[++ne] = (EDGE){v,h[u],f}; h[u] = ne;
	ed[++ne] = (EDGE){u,h[v],0}; h[v] = ne;
}
int cur[maxn],vis[maxn],used[maxn],d[maxn],now,S,T;
int q[maxn],head,tail;
inline bool bfs(){
	q[head = tail = 0] = S; vis[S] = now; d[S] = 0;
	int u;
	while (head <= tail){
		u = q[head++];
		Redge(u) if (ed[k].f && vis[to = ed[k].to] != now){
			d[to] = d[u] + 1; vis[to] = now;
			if (to == T) return true;
			q[++tail] = to;
		}
	}
	return vis[T] == now;
}
int dfs(int u,int minf){
	if (u == T || !minf) return minf;
	int flow = 0,f,to;
	if (used[u] != now) cur[u] = h[u],used[u] = now;
	for (int& k = cur[u]; k; k = ed[k].nxt)
		if (vis[to = ed[k].to] == now && d[to] == d[u] + 1 && (f = dfs(to,min(minf,ed[k].f)))){
			ed[k].f -= f; ed[k ^ 1].f += f;
			flow += f; minf -= f;
			if (!minf) break;
		}
	return flow;
}
int maxflow(){
	int flow = 0; now = 1;
	while (bfs()){
		flow += dfs(S,INF);
		now++;
	}
	return flow;
}
int n,m,w[105][105],x[105],b[maxn],tot,id[105][105],cnt,ans;
int main(){
	n = read(); m = read(); S = 0;
	REP(i,n) b[i] = x[i] = read();
	sort(b + 1,b + 1 + n); tot = 1;
	for (int i = 2; i <= n; i++) if (b[i] != b[tot]) b[++tot] = b[i];
	for (int i = 1; i <= n; i++) x[i] = lower_bound(b + 1,b + 1 + tot,x[i]) - b;
	cnt = n + tot;
	for (int i = 1; i <= n; i++){
		for (int j = i; j <= n; j++){
			id[i][j] = ++cnt;
			w[i][j] = read();
		}
	}
	T = ++cnt;
	REP(i,n){
		build(i,x[i] + n,INF);
		build(i,T,b[x[i]]);
	}
	REP(i,tot) build(i + n,T,m * b[i] * b[i]);
	for (int i = 1; i <= n; i++){
		for (int j = i; j <= n; j++){
			if (w[i][j] >= 0) build(S,id[i][j],w[i][j]),ans += w[i][j];
			else build(id[i][j],T,-w[i][j]);
			for (int k = i; k <= j; k++)
				build(id[i][j],k,INF);
			if (i < j){
				build(id[i][j],id[i + 1][j],INF);
				build(id[i][j],id[i][j - 1],INF);
			}
		}
	}
	printf("%d\n",ans - maxflow());
	return 0;
}