BZOJ 2412: 电路检修

Description

[0,x]中全是1,其余全是0,每个点有一个权值,求最坏情况下得到x的最小权值.

Sol

DP+单调队列.

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http://www.cnblogs.com/beiyuoi/p/5974574.html

Code

/**************************************************************
    Problem: 2448
    User: BeiYu
    Language: C++
    Result: Accepted
    Time:1316 ms
    Memory:48420 kb
****************************************************************/
#include<cstdio>
#include<iostream>
using namespace std;

const int N = 2005;
#define A(x) (f[i][x-1]+a[x])
#define B(x) (f[x+1][j]+a[x])

int n,a[N];
int f[N][N],g[N][N];
int q[N][N],h[N],t[N];

inline int in(int x=0,char ch=getchar(),int v=1){
	while(ch!='-' && (ch>'9'||ch<'0')) ch=getchar();if(ch=='-') v=-1,ch=getchar();
	while(ch>='0' && ch<='9') x=(x<<3)+(x<<1)+ch-'0',ch=getchar();return x*v; }

int main(){
//	freopen("in.in","r",stdin);
	n=in();
	for(int i=1;i<=n;i++) a[i]=in();

	for(int i=n;i;--i){
		f[i][i]=a[i],g[i][i]=i;

		//f[i][j]=min{ f[i][k-1]+t[k] },g[i][j]<=k<=j;  =>q[0]
		//f[i][j]=min{ f[k+1][j]+t[k] },i<=k<g[i][j];  =>q[j]

		h[0]=1,t[0]=0;
		h[i]=1,t[i]=0;
		q[i][++t[i]]=i;

		for(int j=i+1;j<=n;++j){
			//g[i][j]
			g[i][j]=g[i][j-1];
			while(g[i][j]<j && f[i][g[i][j]-1] < f[g[i][j]+1][j]) ++g[i][j];

			//q[0].pop g[i][j-1]--(g[i][j]-1)
			for(int k=g[i][j-1];k<g[i][j];++k)
				if(q[0][h[0]] == k) ++h[0];
			//j->q[0]
			while(h[0]<=t[0] && A(q[0][t[0]]) > A(j)) --t[0];
			q[0][++t[0]]=j;

			//q[j].pop g[i+1][j]-g[i][j]
			for(int k=g[i+1][j];k>=g[i][j];--k)
				if(q[j][h[j]] == k) ++h[j];
			//i->q[j]
			while(h[j]<=t[j] && B(q[j][t[j]]) > B(i)) --t[j];
			q[j][++t[j]]=i;

			//f[i][j]
			f[i][j]=min(A(q[0][h[0]]),B(q[j][h[j]]));

		}
	}

//	for(int i=1;i<=n;i++) for(int j=1;j<=n-i+1;j++) printf("%d%c",g[j][j+i-1]," \n"[j==n-i+1]);
//	cout<<"***"<<endl;
//	for(int i=1;i<=n;i++) for(int j=1;j<=n-i+1;j++) printf("%d%c",f[j][j+i-1]," \n"[j==n-i+1]);

	cout<<f[1][n]<<endl;
	return 0;
}