codeforces 722C (并查集)

题目链接:http://codeforces.com/contest/722/problem/C

题意:每次破坏一个数,求每次操作后的最大连续子串和。

思路:并查集逆向操作

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e5 + 10;
ll sum[N], ans[N];
int n, a[N], b[N], father[N], r[N];
bool vis[N];
int finds(int x)
{
if(father[x] != x)
father[x] = finds(father[x]);
return father[x];
}
void connect(int a,int b)
{
a = finds(a);
b = finds(b);
if(r[a] > r[b])
father[b] = a;
else if(r[a] < r[b])
father[a] = b;
else
{
father[a] = b;
r[b]++;
}
sum[a] = sum[b] = sum[a] + sum[b];
}
void init()
{
for(int i = 1; i <= n + 1; i++)
father[i] = i;
vis[0] = vis[n+1] = 1;
}
int main()
{
scanf("%d",&n);
for(int i = 1; i <= n; i++)
scanf("%d",a+i);
for(int i = 1; i <= n; i++)
scanf("%d",b+i);
init();
for(int i = n; i > 1; i--)
{
sum[b[i]] = a[b[i]];
if(vis[b[i]-1])
connect(b[i] - 1,b[i]);
if(vis[b[i]+1])
connect(b[i] + 1,b[i]);
ans[i-1] = max(ans[i],sum[finds(b[i])]);
vis[b[i]] = 1;
}
for(int i = 1; i <= n; i++)
printf("%I64d\n",ans[i]);
return 0;
}
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