codeforces 722C （并查集）

来源：互联网发布：华云数据集团有限公司编辑：程序博客网时间：2024/04/30 17:26

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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