CF802C Heidi and Library (hard)

题解

买进书有点麻烦..

我们考虑每天强制买进书

那么把书留到这一天,那么就等于在前一天卖了后,这一天又买回来,

卖出的价值为(-C_{a[i]})

然后对每一天建两个点(A,B)

到了(B)点,就等于卖了这本书

然后(A_i->A_{i+1})连((k-1,0))

因为强制买书,所以只有(k-1)本书流向这天

(S->A_{i})连((1,C_{a[i] }))

(A_{i}->B_{i})连((1, 0))

(B_i->T)连((1,0))

记(pos[i])表示书(i)上一次出现的位置

那么(A_{i-1}->B_{pos[i]})连((1,-C_{a[i]}))

然后跑最小费用最大流

Code

#include<bits/stdc++.h>

#define LL long long
#define RG register

using namespace std;
template<class T> inline void read(T &x) {
	x = 0; RG char c = getchar(); bool f = 0;
	while (c != '-' && (c < '0' || c > '9')) c = getchar(); if (c == '-') c = getchar(), f = 1;
	while (c >= '0' && c <= '9') x = x*10+c-48, c = getchar();
	x = f ? -x : x;
	return ;
}
template<class T> inline void write(T x) {
	if (!x) {putchar(48);return ;}
	if (x < 0) x = -x, putchar('-');
	int len = -1, z[20]; while (x > 0) z[++len] = x%10, x /= 10;
	for (RG int i = len; i >= 0; i--) putchar(z[i]+48);return ;
}

const int N = 210, M = 2000010, inf = 2147483647;

struct node {
	int to, nxt, w, v;
}g[M];
int last[N], gl = 1;
void add(int x, int y, int w, int v) {
	g[++gl] = (node) {y, last[x], w, v};
	last[x] = gl;
	g[++gl] = (node) {x, last[y], 0, -v};
	last[y] = gl;
}

int dis[N], s, t, pre[N], from[N];
bool vis[N];

queue<int> q;

inline bool spfa() {
	memset(dis, 127 / 3, sizeof(dis));
	dis[s] = 0;
	q.push(s);
	while (!q.empty()) {
		int u = q.front(); q.pop();
		for (int i = last[u]; i; i = g[i].nxt) {
			int v = g[i].to;
			if (dis[v] > dis[u] + g[i].v && g[i].w) {
				dis[v] = dis[u] + g[i].v;
				pre[v] = u; from[v] = i;
				if (!vis[v]) vis[v] = 1, q.push(v);
			}
		}
		vis[u] = 0;
	}
	return dis[0] != dis[t];
}

int a[85], c[85], n, k;

inline int Mcmf() {
	int res = 0;
	while (spfa()) {
		int di = inf;
	   	for (int i = t; i != s; i = pre[i]) di = min(di, g[from[i]].w);
		res += di * dis[t];
		for (int i = t; i != s; i = pre[i]) g[from[i]].w -= di, g[from[i] ^ 1].w += di;
	}
	return res;
}

int pos[85];

int main() {
	read(n), read(k);
	s = n * 2 + 1, t = s + 1;
	for (int i = 1; i <= n; i++) read(a[i]);
	for (int i = 1; i <= n; i++) read(c[i]);
	for (int i = 1; i <= n; i++) {
		add(s, i, 1, c[a[i]]);
		add(i + n, t, 1, 0);
		add(i, i + n, 1, 0);
		if (pos[a[i]]) add(i - 1, pos[a[i]] + n, 1, -c[a[i]]);
		pos[a[i]] = i;
	}
	for (int i = 1; i < n; i++)
		add(i, i + 1, k - 1, 0);
	printf("%d
", Mcmf());
	return 0;
}