hdoj 5289 线段树

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hdoj 5289

题意：给n个数字，问有多少区间[i, j]（i <= j）满足区间中最大数与最小数差值小于k。

思路：从左起依次寻找能到达的最右端的区间，答案累加 r - l。

#include <cstdio>#include <cstring>const int M = 100005;inline int max(int a, int b){    return a > b ? a : b;}inline int min(int a, int b){    return a < b ? a : b;}#define MAX true#define MIN falsestruct Tree{    int l, r, maxval, minval;}tree[M * 4];int nums[M], n, k;void pullup(int rt){    tree[rt].maxval = max(tree[rt * 2].maxval, tree[rt * 2 + 1].maxval);    tree[rt].minval = min(tree[rt * 2].minval, tree[rt * 2 + 1].minval);}void buildtree(int rt, int l, int r) {    tree[rt].l = l, tree[rt].r = r;    if(l == r){        tree[rt].maxval = tree[rt].minval = nums[l];        return ;    }    int mid = (l + r) / 2;    buildtree(rt * 2, l, mid);    buildtree(rt * 2 + 1, mid + 1, r);    pullup(rt);}int query(int rt, int l, int r, bool flag){    if(tree[rt].l == l && tree[rt].r == r) {        return flag ? tree[rt].maxval : tree[rt].minval;    }    int mid = (tree[rt].l + tree[rt].r) / 2;    if(r <= mid) return query(rt * 2, l, r, flag);    else if(l > mid) return query(rt * 2 + 1, l, r, flag);    else {        if(flag) return max(query(rt * 2, l, mid, flag), query(rt * 2 + 1, mid + 1, r, flag));        else return min(query(rt * 2, l, mid, flag), query(rt * 2 + 1, mid + 1, r, flag));    }}int Scan()     //输入外挂{    int res=0,ch,flag=0;    if((ch=getchar())=='-')        flag=1;    else if(ch>='0'&&ch<='9')        res=ch-'0';    while((ch=getchar())>='0'&&ch<='9')        res=res*10+ch-'0';    return flag?-res:res;}main() {    int t;  //  freopen("in.txt", "r", stdin);    scanf("%d", &t);    while(t--){        memset(tree, 0, sizeof tree);        scanf("%d %d", &n, &k);        for(int i = 1; i <= n; i++) nums[i] = Scan();        nums[n + 1] = 0x7fffffff;        n++;        buildtree(1, 1, n);        int i = 1, j = 1, minval = nums[1], maxval = nums[1];        long long ans = 0;        while(true){            while(maxval - minval < k && j < n) {                j++;                minval = min(minval, nums[j]);                maxval = max(maxval, nums[j]);            }            ans += j - i;            i++;            if(i == n) break;            maxval = query(1, i, j, MAX);            minval = query(1, i, j, MIN);        }        printf("%I64d\n", ans);    }}

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