bzoj 4873: [Shoi2017]寿司餐厅 [最小割]

4873: [Shoi2017]寿司餐厅

题意：略

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唯一会做的...

一眼最小割

就是最大权闭合子图呀

\(s\rightarrow d_{positive} \rightarrow -d_{negtive} \rightarrow t\)

然后区间包含关系连inf

然后向t连花费

一开始看成\(mx^2 + cx\) x是选择种类数，直接吓哭了 平方怎么割啊我只会费用流

然后发现x是编号gg

然后建图注意编号最大是1000，tle了两次....

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long ll;
const int N = 2e4+5, M = 4e6+5, INF = 1e9;
inline int read() {
    char c=getchar(); int x=0,f=1;
    while(c<'0' || c>'9') {if(c=='-')f=-1; c=getchar();}
    while(c>='0' && c<='9') {x=x*10+c-'0'; c=getchar();}
    return x*f;
}

int n, m, a[105], d[105][105], s, t, sum, mx;
struct edge{int v, ne, c, f;} e[M];
int cnt=1, h[N];
inline void ins(int u, int v, int c) { //printf("ins (%d, %d) %d\n", u, v, c);
	e[++cnt] = (edge){v, h[u], c, 0}; h[u] = cnt;
	e[++cnt] = (edge){u, h[v], 0, 0}; h[v] = cnt;
}

namespace mf {
	int vis[N], d[N], q[N], head, tail;
	bool bfs() {
		memset(vis, 0, sizeof(vis));
		head = tail = 1;
		q[tail++] = s; d[s] = 0; vis[s] = 1;
		while(head != tail) {
			int u = q[head++];
			for(int i=h[u];i;i=e[i].ne) 
				if(!vis[e[i].v] && e[i].c > e[i].f) {
					int v = e[i].v;
					d[v] = d[u]+1; vis[v] = 1;
					q[tail++] = v;
					if(v == t) return true;
				}
		}
		return false;
	}
	int cur[N];
	int dfs(int u, int a) {
		if(u == t || a == 0) return a;
		int flow = 0, f;
		for(int &i=cur[u];i;i=e[i].ne)
			if(d[e[i].v] == d[u]+1 && (f = dfs(e[i].v, min(a, e[i].c - e[i].f))) > 0) {
				flow += f;
				e[i].f += f;
				e[i^1].f -= f;
				a -= f;
				if(a == 0) break;
			}
		if(a) d[u] = -1;
		return flow;
	}
	int dinic() {
		int flow = 0;
		while(bfs()) {
			for(int i=s; i<=t; i++) cur[i] = h[i];
			flow += dfs(s, INF);
		}
		return flow;
	}
}

inline int id(int i, int j) {return i==j ? i : i*n+j;}
int vis[N];
void build() {
	int n2 = n*n, n3 = n+n2;
	s = 0; t = n + n2 + mx + 1;
	for(int i=1; i<=n; i++) {
		if(d[i][i] > 0) ins(s, i, d[i][i]);
		else ins(i, t, -d[i][i]);
		ins(i, t, a[i]);
		if(m == 0) continue;
		ins(i, n3 + a[i], INF);
		if(!vis[a[i]]) ins(n3 + a[i], t, a[i] * a[i]), vis[a[i]] = 1;
	}
	for(int i=n; i>=1; i--)
		for(int j=i+1; j<=n; j++) {
			int now = id(i, j);
			if(d[i][j] > 0) ins(s, now, d[i][j]);
			else {ins(now, t, -d[i][j]); continue;}
			int rr = 0;
			for(int l=i; l<=j; l++) {
				int _ = max(i, rr);
				for(int r= l==i ? j-1 : j; r>=_; r--) {
					ins(now, id(l, r), INF);
					if(d[l][r] > 0) { rr = max(rr, r); break; }
				}
			}
		}
}

int main() {
	freopen("in", "r", stdin);
	n=read(); m=read();
	for(int i=1; i<=n; i++) a[i] = read(), mx = max(mx, a[i]);
	for(int i=1; i<=n; i++)
		for(int j=i; j<=n; j++) {
			d[i][j] = read();
			if(d[i][j] > 0) sum += d[i][j];
		}
	build();
	int ans = mf::dinic(); //printf("ans %d\n", ans);
	printf("%d\n", sum - ans);
}