poj-3264

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// 2404K    3688MS  G++#include <stdio.h>#include <string.h>const int MAX = 50010;struct TreeNode {    int left;    int right;    int Max;    int Min;};typedef struct TreeNode TreeNode;TreeNode tree[MAX<<2];#define INF 9999999void buildTree(int pos, int left, int right) {    TreeNode & curNode = tree[pos];    curNode.left = left;    curNode.right = right;    curNode.Max = -INF;    curNode.Min = INF;    if (left == right) {        return;    } else {        int mid = (left + right)>>1;        buildTree(pos<<1, left, mid);        buildTree(pos<<1|1, mid+1, right);    }}void pushUp(int pos) {    if (pos == 1) {        return;    } else {        int parentPos = pos>>1;        TreeNode & parentNode = tree[parentPos];        TreeNode & leftNode = tree[parentPos<<1];        TreeNode & rightNode = tree[parentPos<<1|1];        int childMax = leftNode.Max > rightNode.Max ? leftNode.Max: rightNode.Max;        int childMin = leftNode.Min < rightNode.Min ? leftNode.Min: rightNode.Min;        char continuePushUp = 0;        int parentMax = parentNode.Max;        int parentMin = parentNode.Min;        if (parentMax < childMax) {            continuePushUp = 1;            parentNode.Max = childMax;        }        if (parentMin > childMin) {            continuePushUp = 1;            parentNode.Min = childMin;        }        // if (continuePushUp) {        //     pushUp(parentPos);        // }    }}void updateSingleCow(int pos, int cowId, int cowHeight) {    TreeNode & curNode = tree[pos];    int curLeft = curNode.left;    int curRight = curNode.right;    if (curLeft == curRight && curLeft == cowId) {        curNode.Max = cowHeight;        curNode.Min = cowHeight;    } else {        int mid = (curLeft + curRight)>>1;        if (cowId <= mid) {            updateSingleCow(pos<<1, cowId, cowHeight);        } else {            updateSingleCow(pos<<1|1, cowId, cowHeight);        }    }    pushUp(pos);}void query(int pos, int rangeLeft, int rangeRight, int & max, int & min) {    TreeNode & curNode = tree[pos];    int curLeft = curNode.left;    int curRight = curNode.right;    if (curLeft == rangeLeft && curRight == rangeRight) {        max = curNode.Max;        min = curNode.Min;    } else {        int mid = (curLeft + curRight)>>1;        if (rangeRight <= mid) {            query(pos<<1, rangeLeft, rangeRight, max, min);        } else if (rangeLeft <= mid && rangeRight > mid) {            int max1;            int max2;            int min1;            int min2;            query(pos<<1, rangeLeft, mid, max1, min1);            query(pos<<1|1, mid+1, rangeRight, max2, min2);            max = max1 > max2 ? max1: max2;            min = min1 < min2 ? min1: min2;        } else if (rangeLeft > mid) {            query(pos<<1|1, rangeLeft, rangeRight, max, min);        }    }}int cowNum;int queryNum;int main() {    while (scanf("%d %d", &cowNum, &queryNum) != EOF) {        buildTree(1, 1, cowNum);        for (int i = 1; i <= cowNum; i++) {            int cowHeight;            scanf("%d", &cowHeight);            updateSingleCow(1, i, cowHeight);        }        int rangeBegin;        int rangeEnd;        for (int i = 1; i <= queryNum; i++) {            int max;            int min;            scanf("%d %d", &rangeBegin, &rangeEnd);            query(1, rangeBegin, rangeEnd, max, min);            printf("%d\n", max - min);        }    }}

险过... 果然这道题用线段树做不是最优,应该有专门的RMQ算法,只是手痒,用线段树搞了一把,

WA了一次,因为给的INF一开始是99999,而题目正常值最大能到100000 ......

从线段树角度看,是到很简单的题,每个节点保存当前区间的最大和最小值即可,查询时根据区间分布情况,递归或者直接比较即可。

这道题其实没有最充分的利用线段树的优点: 可以适应动态变化(一开始初始化cow的高度也算是种动态变化吧) 以及 延迟标记, 毕竟不是为线段树而生的题.


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