XVII Open Cup named after E.V. Pankratiev Grand Prix of Moscow Workshops, Sunday, April 23, 2017 Problem D. Great Again

题目：

Problem D. Great Again
Input file: standard input
Output file: standard output
Time limit: 2 seconds
Memory limit: 512 megabytes
The election in Berland is coming. The party United Berland is going to use its influence to win them
again. The crucial condition for the party is to win the election in the capital to show the world that the
protests of opposition in it are inspired by external enemies.
The capital of Berland consists of only one long road with n people living alongside it. United Berland
has a lot of informers, so they know for each citizen whether he is going to attend the election, and if yes,
who is he going to vote for: the ruling party or the opposition.
United Berland has a vast soft power, so they can lobby the desired distribution of districts. Every district
should be a consecutive segment of the road of length between l and r inclusive. Each citizen must be
assigned to exactly one district. The votes are counted in each district separately, and the parties receive
one point for each district, where it receives strictly more votes than the other party. If the parties got
equal result in this district, no one gets its vote. United Berland is going to create the distribution that
maximizes the difference of its points and points of the opposition, and you are asked to compute this
value.
Input
The first line of the input contains three positive integers n, l, r (1 ≤ n ≤ 300 000, 1 ≤ l ≤ r ≤ n) — the
number of citizens in the capital, the lower and the upper bounds on the possible length of a district.
The second line contains n integers a1; a2; : : : ; an (ai 2 f-1; 0; 1g), denoting the votes of the citizens. 1
means vote for the ruling party, -1 means vote for opposition, 0 means that this citizen is not going to
come to the elections.
Output
If there is no way to divide the road into districts of lengths between l and r, print “Impossible” (without
quotes).
Otherwise, print one integer — the maximum possible difference between the scores of United Berland
and the opposition in a valid distribution of citizens among voting districts.
Examples

standard input	standard output
5 1 5 1 -1 0 -1 1	1
5 2 3 -1 1 -1 1 -1	-1
6 1 1 1 -1 -1 -1 -1 -1	-4
5 3 3 1 1 1 1 1	Impossible

Note
In the first sample, the optimal division of districts is f1g; f2; 3; 4g; f5g.
In the second sample, the optimal division is f1; 2g; f3; 4; 5g.
In the third sample, there is only one possible division.
There is no way to divide 5 in segments of length 3, so in the fourth sample the answer is “Impossible”.

思路：

　　DP：dp[i]=max{dp[j]+f[j+1][i]},（i-l+1<=j<=l-r+1）

　　现在难点是怎么做到快速转移。（f[j+1][i]表示区间[j+1,i]的贡献）

#include <bits/stdc++.h>

using namespace std;

#define MP make_pair
#define PB push_back
typedef long long LL;
typedef pair<int,int> PII;
const double eps=1e-8;
const double pi=acos(-1.0);
const int K=3e5+7;
const int mod=1e9+7;

int n,tl,tr,py,sum[K],dp[K],v[K*20];
priority_queue<PII>q[K*4];
int update(int o,int l,int r,int pos,int x)
{
    if(l==r) return v[o]=x;
    int mid=l+r>>1;
    if(pos<=mid) update(o<<1,l,mid,pos,x);
    else update(o<<1|1,mid+1,r,pos,x);
    v[o]=max(v[o<<1],v[o<<1|1]);
}
int query(int o,int l,int r,int nl,int nr)
{
    if(l==nl&&r==nr) return v[o];
    int mid=l+r>>1;
    if(nr<=mid) return query(o<<1,l,mid,nl,nr);
    else if(nl>mid) return query(o<<1|1,mid+1,r,mid+1,nr);
    return max(query(o<<1,l,mid,nl,mid),query(o<<1|1,mid+1,r,mid+1,nr));
}
void add(int x)
{
    if(x<0) return ;
    int fx=sum[x]+n+2;
    if(q[fx].size()==0||q[fx].top().first<dp[x])
        update(1,1,2*n+2,fx,dp[x]);
    q[fx].push(MP(dp[x],x));
}
void del(int x)
{
    if(x<0) return;
    int fx=sum[x]+n+2;
    while(q[fx].size()&&q[fx].top().second<=x) q[fx].pop();
    if(q[fx].size()==0)
        update(1,1,2*n+2,fx,-mod);
    else
        update(1,1,2*n+2,fx,q[fx].top().first);
}
int main(void)
{
    scanf("%d%d%d",&n,&tl,&tr);
    for(int i=1,mx=n*8+2;i<=mx;i++) v[i]=-mod;
    for(int i=1,x;i<=n;i++) scanf("%d",&x),sum[i]=sum[i-1]+x;
    for(int i=1;i<=n;i++)
    {
        del(i-tr-1);add(i-tl);
        int q1=query(1,1,2*n+2,1,sum[i]-1+n+2);
        int q2=query(1,1,2*n+2,sum[i]+n+2,sum[i]+n+2);
        int q3=query(1,1,2*n+2,sum[i]+1+n+2,n+n+2);
        if(q1==q2&&q2==q3&&q1==-mod)
            dp[i]=-mod;
        else
            dp[i]=max(max(q1+1,q2),q3-1);
    }
    if(dp[n]==-mod) printf("Impossible\n");
    else printf("%d",dp[n]);
    return 0;
}