CF 818 G Four Melodies 题解

CF 818 G Four Melodies 题解

根据题目要求连边，跑最小费用流。

一个优化技巧：由于spfa会退化成(O(n imes m)),所以需要利用dijkstra+势函数来优化。第一次最短路可以dp出来，由于是DAG。

/*
{
######################
#       Author       #
#        Gary        #
#        2020        #
######################
*/
#include<bits/stdc++.h>
#define rb(a,b,c) for(int a=b;a<=c;++a)
#define rl(a,b,c) for(int a=b;a>=c;--a)
#define LL long long
#define IT iterator
#define PB push_back
#define II(a,b) make_pair(a,b)
#define FIR first
#define SEC second
#define FREO freopen("check.out","w",stdout)
#define rep(a,b) for(int a=0;a<b;++a)
#define SRAND mt19937 rng(chrono::steady_clock::now().time_since_epoch().count())
#define random(a) rng()%a
#define ALL(a) a.begin(),a.end()
#define POB pop_back
#define ff fflush(stdout)
#define fastio ios::sync_with_stdio(false)
#define check_min(a,b) a=min(a,b)
#define check_max(a,b) a=max(a,b)
using namespace std;
//inline int read(){
//    int x=0;
//    char ch=getchar();
//    while(ch<'0'||ch>'9'){
//        ch=getchar();
//    }
//    while(ch>='0'&&ch<='9'){
//        x=(x<<1)+(x<<3)+(ch^48);
//        ch=getchar();
//    }
//    return x;
//}
const int INF=0x3f3f3f3f;
typedef pair<int,int> mp;
/*}
*/
int a[4004];
int n;

const int GRAPH_SIZE= 6000+20;
int s=0,t=GRAPH_SIZE-1;
struct EDGE{
	int u,v,c,cos;
};
vector<EDGE> e;
vector<int> each[GRAPH_SIZE];
int maxflow,mincost;
int flow[GRAPH_SIZE];
int dis[GRAPH_SIZE],inq[GRAPH_SIZE],las[GRAPH_SIZE];
bool spfa(){
	memset(inq,0,sizeof(inq));
	memset(dis,63,sizeof(dis));
	flow[s]=INF;
	dis[s]=0;
	queue<int> q;
	q.push(s);
	inq[s]=1;
	while(!q.empty()){
		int now=q.front();
		q.pop();
		inq[now]=0;
		for(auto it:each[now]){
			int to,f,c;
			to=e[it].v;
			f=e[it].c;
			c=e[it].cos;
			if(f<=0) continue;
			if(dis[to]>dis[now]+c){
				dis[to]=dis[now]+c;
				las[to]=it;
				flow[to]=min(flow[now],f);
				if(!inq[to]) q.push(to);
				inq[to]=1;		
			}
		}
	}
	return dis[t]!=INF;
}
void KM(){
	while(spfa()){
		maxflow+=flow[t];
		mincost+=dis[t]*flow[t];
		int now=t;
		while(now!=s){
			e[las[now]].c-=flow[t];
			e[las[now]^1].c+=flow[t];
			now=e[las[now]].u;
		}
	}
}
void make_edge(int U,int V,int C,int COS){
	EDGE tmp;
	tmp.u=U;
	tmp.v=V;
	tmp.c=C;
	tmp.cos=COS;
	e.PB(tmp);
	each[U].PB(e.size()-1);
	swap(tmp.u,tmp.v);
	tmp.c=0;
	tmp.cos=-COS;
	e.PB(tmp);
	each[V].PB(e.size()-1);
}
int lasval[100000+20];
int main(){
	scanf("%d",&n);
	rb(i,1,n) scanf("%d",&a[i]);
	rb(i,1,n) make_edge(2*i-1,2*i,1,-1);
	int las[7]={0};
	rb(i,1,n){
		if(las[a[i]%7]){
			make_edge(2*las[a[i]%7]-1,2*i-1,INF,0);
			make_edge(2*las[a[i]%7],2*i-1,INF,0);
		}
		if(lasval[a[i]]){
			make_edge(2*lasval[a[i]]-1,2*i-1,INF,0);
		}
		if(lasval[a[i]-1]){
			make_edge(2*lasval[a[i]-1],2*i-1,INF,0);
		}
		if(lasval[a[i]+1]){
			make_edge(2*lasval[a[i]+1],2*i-1,INF,0);
		}
		las[a[i]%7]=lasval[a[i]]=i; 
	}
	int supers=2*n+1;
	rb(i,1,n) make_edge(supers,2*i-1,1,0),make_edge(2*i,t,1,0);
	make_edge(s,supers,4,0);
	KM();
	printf("%d
",-mincost);
	return 0;
}