数值标记问题 离线+树状数组 HDU 3938 + HDU 3333

HDU 3938

题目大意：给你一个长度为n的数组a，定义区间[l,r]的val为区间内所有不同的数值之和。现在有m个询问，每次询问一个区间，问区间的val是多少。

思路：将所有的询问按照右端点排序。然后暴力枚举右区间，然后对之前出现过的val做一个标记即可，每次都更新这个标记就好了。 具体的和HDU 5869一样，只不过5869还要预处理，比较难

//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
const int maxn = 50000 + 5;
const int maxm = 200000 + 5;
const int maxval = 1000000 + 5;
LL tree[maxn], a[maxn], ans[maxm];
vector<pair<int, int> > q[maxn];
int n, m;
int pre[maxval];
inline int lowbit(int x) {return x & -x;}

void update(int x, int val){
    for (int i = x; i <= n; i += lowbit(i))
        tree[i] += val;
}

LL sum(int x){
    LL ans = 0;
    for (int i = x; i > 0; i -= lowbit(i)){
        ans += tree[i];
    }
    return ans;
}

int main(){
    int t; cin >> t;
    while (t--){
        memset(pre, 0, sizeof(pre));
        memset(tree, 0, sizeof(tree));
        scanf("%d", &n);
        for (int i = 1; i <= n; i++) {
            scanf("%I64d", a + i);
            q[i].clear();
        }
        scanf("%d", &m);
        for (int i = 1; i <= m; i++){
            int l, r; scanf("%d%d", &l ,&r);
            q[r].pb(mk(l, i));
        }
        for (int i = 1; i <= n; i++){
            if (pre[a[i]]){
                update(pre[a[i]], -1 * a[i]);
            }
            pre[a[i]] = i;
            update(i, a[i]);
            int len = q[i].size();
            for (int j = 0; j < len; j++){
                pair<int, int> p = q[i][j];
                ans[p.second] = sum(i) - sum(p.first - 1);
            }
        }
        for (int i = 1; i <= m; i++){
            printf("%I64d
", ans[i]);
        }
    }
    return 0;
}

HDU 3333

这道题就是a数组的范围大了一点而已，没啥的，只要离散化一下就好了，其他的思路都一样

然后代码只需要多添加一个离散化即可

//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
const int maxn = 30000 + 5;
const int maxm = 100000 + 5;
LL tree[maxn], a[maxn], b[maxn], ans[maxm];
vector<pair<int, int> > q[maxn];
int n, m;
int pre[maxn];
inline int lowbit(int x) {return x & -x;}

void update(int x, int val){
    for (int i = x; i <= n; i += lowbit(i))
        tree[i] += val;
}

LL sum(int x){
    LL ans = 0;
    for (int i = x; i > 0; i -= lowbit(i)){
        ans += tree[i];
    }
    return ans;
}

int main(){
    int t; cin >> t;
    while (t--){
        memset(pre, 0, sizeof(pre));
        memset(tree, 0, sizeof(tree));
        scanf("%d", &n);
        for (int i = 1; i <= n; i++) {
            scanf("%I64d", a + i);
            b[i] = a[i];
            q[i].clear();
        }
        sort(b + 1, b + 1 + n);
        scanf("%d", &m);
        for (int i = 1; i <= m; i++){
            int l, r; scanf("%d%d", &l ,&r);
            q[r].pb(mk(l, i));
        }
        for (int i = 1; i <= n; i++){
            int posa = lower_bound(b + 1, b + 1 + n, a[i]) - b;
            if (pre[posa]){
                update(pre[posa], -1 * a[i]);
            }
            pre[posa] = i;
            update(i, a[i]);
            int len = q[i].size();
            for (int j = 0; j < len; j++){
                pair<int, int> p = q[i][j];
                ans[p.second] = sum(i) - sum(p.first - 1);
            }
        }
        for (int i = 1; i <= m; i++){
            printf("%I64d
", ans[i]);
        }
    }
    return 0;
}