codeforces 并查集_codeforces 722C （并查集）

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

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("%I64dn",ans[i]);

return 0;

}