【牛客网】Idol Master

【牛客网】Idol Master

也是一道网络流解线性规划

不过需要从小于号的那边解

限制是\(a \leq \sum_{i = 1}^{k} x_{i}\leq b\)

其中\(0 \leq x_{i} \leq 1\)

\(\sum_{i = 1}^{k} x_{i} = b - y_{i}\)

\(y_i \leq b - a\)

然后把\(N - k + 1\)个式子都列出来，前后补上0=0，就可以解了

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define ba 47
#define MAXN 200005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
	if(c == '-') f = -1;
	c = getchar();
    }
    while(c >= '0' && c <= '9') {
	res = res * 10 +c - '0';
	c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
	out(x / 10);
    }
    putchar('0' + x % 10);
}
struct node {
    int to,next,cap;int64 val;
}E[1000005];
int head[405],sumE = 1,Ncnt,S,T;
int64 ans,c[405];
int N,K,A,B;
bool vis[405];
void add(int u,int v,int c,int a) {
    E[++sumE].to = v;
    E[sumE].next = head[u];
    E[sumE].cap = c;
    E[sumE].val = a;
    head[u] = sumE;
}
void addtwo(int u,int v,int c,int a) {
    add(u,v,c,a);add(v,u,0,-a);
}
int64 dis[405];
bool inq[405];
queue<int> Q;
bool SPFA() {
    for(int i = 1 ; i <= Ncnt ; ++i) dis[i] = -1e18;
    dis[S] = 0;Q.push(S);
    memset(inq,0,sizeof(inq));
    while(!Q.empty()) {
	int u = Q.front();Q.pop();
	inq[u] = 0;
	for(int i = head[u] ; i ; i = E[i].next) {
	    int v = E[i].to;
	    if(E[i].cap) {
		if(dis[u] + E[i].val > dis[v]) {
		    dis[v] = dis[u] + E[i].val;
		    if(!inq[v]) {Q.push(v);inq[v] = 1;}
		}
	    }
	}
    }
    return dis[T] > -1e18;
}
int dfs(int u,int aug) {
    if(u == T) {
	ans += aug * dis[T];
	return aug;
    }
    vis[u] = 1;
    for(int i = head[u] ; i ; i = E[i].next) {
	int v = E[i].to;
	if(!vis[v] && E[i].cap && dis[v] == dis[u] + E[i].val) {
	    int t = dfs(v,min(E[i].cap,aug));
	    if(t) {
		E[i].cap -= t;
		E[i ^ 1].cap += t;
		return t;
	    }
	}
    }
    return 0;
}
void MCMF() {
    while(SPFA()) {
	do {
	    memset(vis,0,sizeof(vis));
	}while(dfs(S,0x7fffffff));
    }
}
void Solve() {
    ans = 0;memset(head,0,sizeof(head));sumE = 1;
    read(N);read(K);read(A);read(B);
    for(int i = 1 ; i <= N ; ++i) {
	read(c[i]);
    }
    int d = N - K + 2;
    S = d + 1;T = d + 2;Ncnt = d + 2;
    for(int i = 1 ; i <= K ; ++i) {
	addtwo(min(i + 1,d),1,1,c[i]);
    }
    for(int i = K + 1 ; i <= N ; ++i) {
	addtwo(min(i + 1,d),i - K + 1,1,c[i]);
    }
    for(int i = 2 ; i <= d ; ++i) {
	addtwo(i,i - 1,B - A,0);
    }
    addtwo(1,T,B,0);
    addtwo(S,d,B,0);
    MCMF();
    out(ans);enter;
}
int main(){
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    int T;
    read(T);
    for(int i = 1 ; i <= T ; ++i) Solve();
    return 0;
}