Codeforces 272C Dima and Staircase 思维 or 线段树
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Codeforces 272C Dima and Staircase 思维 or 线段树
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題意:給出n個物品高度ai,ai<=1e9,ai非遞減,n,m<=1e5,m個方塊的寬度和高度,問m個方塊依次落下時達到的高度
每次都從左端點1開始疊放,第i個方塊肯定落在第i-1個方塊的上方 取高度邊界low=ans[i-1]+h[i-1]
高度遞增,(下標>=i)高度>=a[i]的都能放的寬度為i的方塊
若l<a[w] ans[i]=a[w]?
若l>=a[w] ans[i]=l (因為高度l以上的寬度都大于等于w)?
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對于每一個箱子,只需要查詢[1,Wi]內的最大值,箱子會被這個最大值支撐著,所以輸出這個最大值,放上這個箱子之后,這個區間內的所有值都會變成上述的那個最大值加上箱子的高度Hi。用線段樹維護即可。
#include<cstdio> #include<cstring> #include<algorithm> #define LL long long #define maxn 100005 using namespace std; struct node {int l, r;LL c, maxh; }tree[maxn << 2]; int h[maxn]; void pushdown(int id) {if (tree[id].c){tree[id << 1].c = tree[id].c;tree[id << 1 | 1].c = tree[id].c;tree[id << 1].maxh = tree[id].maxh;tree[id << 1 | 1].maxh = tree[id].maxh;} } void build(int id, int l, int r) {tree[id].l = l;tree[id].r = r;if (tree[id].l == tree[id].r){tree[id].maxh = h[l];}else{int mid = (tree[id].l + tree[id].r) >> 1;build(id << 1, l, mid);build(id << 1 | 1, mid + 1, r);tree[id].maxh = max(tree[id << 1].maxh, tree[id << 1 | 1].maxh);} } void op(int id, int l, int r, LL c) {if (l <= tree[id].l&&tree[id].r <= r){tree[id].c = c;tree[id].maxh = c;}else{pushdown(id);int mid = (tree[id].l + tree[id].r) >> 1;if (l <= mid) op(id << 1, l, r, c);if (mid<r) op(id << 1 | 1, l, r, c);tree[id].maxh = max(tree[id << 1].maxh, tree[id << 1 | 1].maxh);} } LL que(int id, int l, int r) {if (l <= tree[id].l&&tree[id].r <= r){return tree[id].maxh;}else{pushdown(id);int mid = (tree[id].l + tree[id].r) >> 1;LL res = 0;if (l <= mid) res = max(res, que(id << 1, l, r));if (mid<r) res = max(res, que(id << 1 | 1, l, r));return res;} } int main() {int n;scanf("%d", &n);for (int i = 1; i <= n; i++) scanf("%d", &h[i]);int m;scanf("%d", &m);build(1, 1, n);while (m--){LL w, h;scanf("%I64d %I64d", &w, &h);LL tmp = que(1, 1, w);printf("%I64d\n", tmp);op(1, 1, w, tmp + h);}return 0; }
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