题意描述
You are given an array consisting of n non-negative integers a 1, a 2, …, a n.
You are going to destroy integers in the array one by one. Thus, you are given the permutation of integers from 1 to n defining the order elements of the array are destroyed.
After each element is destroyed you have to find out the segment of the array, such that it contains no destroyed elements and the sum of its elements is maximum possible. The sum of elements in the empty segment is considered to be 0.
每次去掉一个数字，求剩余数字的最大子段和
思路
这道题可以反着来考虑，我们从后往前还是遍历，每次添加一个数字，使用并查集来维护子段和。
AC代码
代码语言：javascript复制#include<iostream>
#include<algorithm>
#include<cstring>
#include<vector>
#define x first
#define y second
#define PB push_back
#define mst(x,a) memset(x,a,sizeof(x))
#define all(a) begin(a),end(a)
#define rep(x,l,u) for(ll x=l;x<u;x )
#define rrep(x,l,u) for(ll x=l;x>=u;x--)
#define IOS ios::sync_with_stdio(false);cin.tie(0);
using namespace std;
typedef unsigned long long ULL;
typedef pair<int,int> PII;
typedef pair<long,long> PLL;
typedef pair<char,char> PCC;
typedef long long ll;
const int N=1e5 10;
const int M=150;
const int INF=0x3f3f3f3f;
const int MOD=998244353;
ll a[N],b[N],ans[N],w[N];
bool used[N];
struct DSU{
vector<int> data;
void init(int n){data.assign(n,-1);}
bool unionSet(int x,int y){
x=root(x);
y=root(y);
if(x!=y){
//if(data[y] < data[x]) swap(x,y);//使用带权并查集时注释掉
data[x] =data[y];
data[y]=x;
}
return x != y;
}
bool same(int x,int y) {return root(x)==root(y);}
int root(int x) {return data[x] < 0 ? x : data[x]=root(data[x]);}
int Size(int x) {return -data[root(x)];}
};
void solve(){
mst(used,false);
int n;cin>>n;
DSU d;
d.init(n 5);
rep(i,1,n 1) cin>>a[i];
rep(i,1,n 1) cin>>b[i];
ans[n]=0;
rrep(i,n,2){
int pos=b[i];
w[pos]=a[pos];//带权并查集
if(used[pos-1]){//前面有数就合并
int x=d.root(pos);
int y=d.root(pos-1);
d.unionSet(y,x);
w[y] =w[x];//权值相加
}
if(used[pos 1]){//后面有数就合并
int x=d.root(pos);
int y=d.root(pos 1);
d.unionSet(y,x);
w[y] =w[x];//权值相加
}
used[pos]=true;
ans[i-1]=max(ans[i],w[d.root(pos)]);//求max
}
rep(i,1,n 1) cout<<ans[i]<<endl;
}
int main(){
IOS;
solve();
return 0;
}
