[HDU 4747 Mex] Mex函数 线段树

·题目

http://acm.hdu.edu.cn/showproblem.php?pid=4747

·分析

先预处理出mex[1][i]，记录每个位置下次出现的位置next[i]，这里大于n的数都可以看作n+1

然后从前往后扫描，对于同一个i，容易发现mex[i][j]是随j递增的

于是找到i和next[i]-1之间第一个mex[i][j]比a[i]大的位置x

那么mex[i+1][x]到mex[i+1][next[i]-1]之间的值一定都为a[i]，因为这一段a[i]空缺

于是转化为区间求和和区间赋值

·代码

/************************************************** *        Problem:  HDU 4747 *         Author:  clavichord93 *          State:  Accepted **************************************************/#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>using namespace std;typedef long long ll;const int MAX_N = 200005;int n;int a[MAX_N];int mex[MAX_N];int last[MAX_N];int next[MAX_N];ll sum[MAX_N << 2];int maxv[MAX_N << 2];int tag[MAX_N << 2];bool vis[MAX_N];#define lch(t) (t << 1)#define rch(t) (t << 1 | 1)void makeTree(int t, int l, int r) {    if (l == r) {        sum[t] = mex[l];        maxv[t] = mex[l];        tag[t] = -1;    }    else {        int mid = (l + r) >> 1;        makeTree(lch(t), l, mid);        makeTree(rch(t), mid + 1, r);        sum[t] = sum[lch(t)] + sum[rch(t)];        maxv[t] = max(maxv[lch(t)], maxv[rch(t)]);        tag[t] = -1;    }}void pushdown(int t, int l, int r) {    if (tag[t] != -1) {        int lt = lch(t);        int rt = rch(t);        int mid = (l + r) >> 1;        sum[lt] = tag[t] * (mid - l + 1);        maxv[lt] = tag[t];        tag[lt] = tag[t];        sum[rt] = tag[t] * (r - mid);        maxv[rt] = tag[t];        tag[rt] = tag[t];        tag[t] = -1;    }}ll getSum(int t, int l, int r, int x, int y) {    if (x <= l && r <= y) {        return sum[t];    }    else {        pushdown(t, l, r);        int mid = (l + r) >> 1;        ll sum = 0;        if (x <= mid) {            sum += getSum(lch(t), l, mid, x, y);        }        if (y > mid) {            sum += getSum(rch(t), mid + 1, r, x, y);        }        return sum;    }}int getPos(int t, int l, int r, int x, int y, int val) {    if (l == r) {        return l;    }    else {        pushdown(t, l, r);        int mid = (l + r) >> 1;        if (x <= mid && maxv[lch(t)] > val) {            return getPos(lch(t), l, mid, x, y, val);        }        if (y > mid && maxv[rch(t)] > val) {            return getPos(rch(t), mid +1, r, x, y, val);        }        return -1;    }}void change(int t, int l, int r, int x, int y, int val) {    if (x <= l && r <= y) {        tag[t] = val;        sum[t] = (ll)(r - l + 1) * val;        maxv[t] = val;    }    else {        pushdown(t, l, r);        int mid = (l + r) >> 1;        if (x <= mid) {            change(lch(t), l, mid, x, y, val);        }        if (y > mid) {            change(rch(t), mid + 1, r, x, y, val);        }        sum[t] = sum[lch(t)] + sum[rch(t)];        maxv[t] = max(maxv[lch(t)], maxv[rch(t)]);    }}#undef lch#undef rchint main() {    #ifdef LOCAL_JUDGE    freopen("in.txt", "r", stdin);    freopen("out.txt", "w", stdout);    #endif    while (scanf("%d", &n), n) {        for (int i = 1; i <= n; i++) {            scanf("%d", &a[i]);            if (a[i] > n) {                a[i] = n + 1;            }        }        for (int i = 0; i <= n + 1; i++) {            last[i] = n + 1;            vis[i] = 0;        }        for (int i = n; i >= 1; i--) {            next[i] = last[a[i]];            last[a[i]] = i;        }        int last = 0;        for (int i = 1; i <= n; i++) {            vis[a[i]] = 1;            while (vis[last]) {                last++;            }            mex[i] = last;        }        //for (int i = 1; i <= n; i++) {            //printf("%d ", mex[i]);        //}        //printf("\n");        makeTree(1, 1, n);        ll ans = 0;        for (int i = 1; i <= n; i++) {            ans += getSum(1, 1, n, i, n);            int x = getPos(1, 1, n, i, next[i] - 1, a[i]);            //cout << "Ans = " << ans << endl;            //cout << "Pos = " << x << endl;            //cout << "Next = " << next[i] << endl;            //cout << "A[i] = " << a[i] << endl;            //cout << endl;            if (1 <= x && x <= next[i] - 1) {                change(1, 1, n, x, next[i] - 1, a[i]);            }        }        printf("%I64d\n", ans);    }    return 0;}