线段树 或者 并查集 Intel Code Challenge Elimination Round (Div.1 + Div.2, combined) C
http://codeforces.com/contest/722/problem/C
题目大意:给你一个串,每次删除串中的一个pos,问剩下的串中,连续的最大和是多少。
思路一:正方向考虑问题,那么就线段树+分类讨论一下就好了,然后代码中flag表示能否转移
//看看会不会爆int!数组会不会少了一维!
//取物问题一定要小心先手胜利的条件
#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define ALL(a) a.begin(), a.end()
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define haha printf("haha\n")
const int maxn = + ;
struct Tree{
bool lf, rf;
bool flag;
LL val, lval, rval;
}tree[maxn << ];
int n;
int b[maxn]; inline void push_up(int o){
int lb = o << , rb = o << | ;
if (tree[lb].flag && tree[rb].flag){
tree[o].lf = tree[o].rf = true;
tree[o].lval = tree[o].rval = tree[o].val = tree[lb].val + tree[rb].val;
return ;
}
tree[o].flag = false; if (tree[lb].lf == false) tree[o].lf = false, tree[o].lval = ;
else {
tree[o].lf = true;
if (tree[lb].flag && tree[rb].lf) tree[o].lval = tree[lb].lval + tree[rb].lval;
else tree[o].lval = tree[lb].lval;
} if (tree[rb].rf == false) tree[o].rf = false, tree[o].rval = ;
else {
tree[o].rf = true;
if (tree[rb].flag && tree[lb].rf) tree[o].rval = tree[rb].rval + tree[lb].rval;
else tree[o].rval = tree[rb].rval;
} tree[o].val = max(tree[lb].val, tree[rb].val);
if (tree[lb].rf && tree[rb].lf){
tree[o].val = max(tree[o].val, tree[lb].rval + tree[rb].lval);
} } void build_tree(int l, int r, int o){
if (l == r) {
LL val;
scanf("%lld", &val);
tree[o].val = tree[o].lval = tree[o].rval = val;
tree[o].flag = tree[o].lf = tree[o].rf = true;
return ;
}
tree[o].lf = tree[o].rf = tree[o].flag = true;
int mid = (l + r) / ;
if (l <= mid) build_tree(l, mid, o << );
if (r > mid) build_tree(mid + , r, o << | );
push_up(o);
} void update(int l, int r, int pos, int o){
if (l == r && l == pos){
tree[o].val = ;
tree[o].flag = tree[o].lf = tree[o].rf = false;
return ;
}
int mid = (l + r) / ;
if (pos <= mid) update(l, mid, pos, o << );
if (pos > mid) update(mid + , r, pos, o << | );
push_up(o);
} int main(){
scanf("%d", &n);
build_tree(, n, ); for (int i = ; i <= n; i++){
int pos; scanf("%d", &pos);
update(, n, pos, );
printf("%lld\n", tree[].val);
}
return ;
}
思路二:逆向考虑,使用并查集,因为最初是0,所以我们只需要逆向考虑就好了。
//看看会不会爆int!数组会不会少了一维!
//取物问题一定要小心先手胜利的条件
#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define ALL(a) a.begin(), a.end()
#define pb push_back
#define mk make_pair
#define fi first
#define se second
#define haha printf("haha\n")
const int maxn = 1e5 + ;
LL a[maxn], b[maxn], sum[maxn], ans[maxn];
bool vis[maxn];
int n;
int par[maxn]; int pfind(int x){
if (x == par[x]) return x;
return par[x] = pfind(par[x]);
} void unite(int x, int y){
x = pfind(x), y = pfind(y);
par[x] = y;
sum[y] += sum[x];
} int main(){
cin >> n;
for (int i = ; i <= n; i++) {
par[i] = i;
scanf("%lld", a + i);
sum[i] = a[i];
}
for (int i = ; i <= n; i++)
scanf("%d", b + i);
LL res = ;
for (int i = n; i >= ; i--){
ans[i] = res;
if (vis[b[i] + ]) unite(b[i], b[i] + );
if (vis[b[i] - ]) unite(b[i], b[i] - );
vis[b[i]] = true;
res = max(res, sum[pfind(b[i])]);
}
for (int i = ; i <= n; i++)
cout << ans[i] << endl;
return ;
}
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