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
",ans[i]);
    return 0;
}