http://acm.hdu.edu.cn/showproblem.php?pid=1779 線段樹 求區間最大值 結點更新

 

#include <iostream><br />#include <cstdio><br />#define LL(x) ((x) << 1)<br />#define RR(x) ((x) << 1 | 1)<br />using namespace std;<br />const int N = 100005;<br />struct Seg_tree {<br />int l, r;<br />int val; //本區間累加值<br />int maxval; //記錄子區間最大值<br />int loction; //子區間最大值位置<br />int mid() {<br />return (l + r) >> 1;<br />}<br />} tree[4 * N];<br />bool readint(int &ret){<br />int sgn;<br />char c;<br />c = getchar();<br />if(c == EOF )<br />return true;<br />while(c != '-' && c < '0' || c > '9')<br />c = getchar();<br />sgn = (c == '-') ? -1 : 1;<br />ret = (c == '-') ? 0 : (c - '0');<br />while((c = getchar()) >= '0' && c <= '9')<br />ret = ret * 10 + (c - '0');<br />ret *= sgn;<br />return false;<br />}<br />void Build(int node, int l, int r) {<br />tree[node].l = l;<br />tree[node].r = r;<br />tree[node].val = 0;<br />tree[node].maxval = 0;<br />tree[node].loction = l;<br />if (l == r) {<br />return;<br />}<br />int mid = (l + r) >> 1;<br />Build(LL(node), l, mid);<br />Build(RR(node), mid + 1, r);<br />}<br />int Update(int node, int l, int r, int val) {<br />if (tree[node].l == l && tree[node].r == r) {<br />tree[node].val += val;<br />return tree[node].maxval + tree[node].val;<br />}<br />int mid = tree[node].mid();<br />if (l <= mid) {<br />if (r <= mid) {<br />int temp = Update(LL(node), l, r, val);<br />if (temp >= tree[RR(node)].maxval + tree[RR(node)].val) {<br />tree[node].maxval = temp;<br />tree[node].loction = tree[LL(node)].loction;<br />} else {<br />tree[node].maxval = tree[RR(node)].maxval + tree[RR(node)].val;<br />tree[node].loction = tree[RR(node)].loction;<br />}<br />} else {<br />tree[node].maxval = Update(LL(node), l, mid, val);<br />tree[node].loction = tree[LL(node)].loction;<br />int temp = Update(RR(node), mid + 1, r, val);<br />if (temp > tree[node].maxval) {<br />tree[node].maxval = temp;<br />tree[node].loction = tree[RR(node)].loction;<br />}<br />}<br />} else {<br />int temp = Update(RR(node), l, r, val);<br />if (temp > tree[LL(node)].val + tree[LL(node)].maxval) {<br />tree[node].maxval = temp;<br />tree[node].loction = tree[RR(node)].loction;<br />} else {<br />tree[node].maxval = tree[LL(node)].val + tree[LL(node)].maxval;<br />tree[node].loction = tree[LL(node)].loction;<br />}<br />}<br />return tree[node].maxval + tree[node].val;<br />}<br />int Query(int node, int l, int r, int &loc) {<br />if (l == tree[node].l && tree[node].r == r) {<br />loc = tree[node].loction;<br />return tree[node].maxval + tree[node].val;<br />}<br />int mid = tree[node].mid();<br />int temp1, temp2;<br />if (l <= mid) {<br />if (r <= mid) {<br />temp1 = Query(LL(node), l, r, loc);<br />} else {<br />int loc2;<br />temp1 = Query(LL(node), l, mid, loc);<br />temp2 = Query(RR(node), mid + 1, r, loc2);<br />if (temp2 > temp1) {<br />temp1 = temp2;<br />loc = loc2;<br />}<br />}<br />} else {<br />temp1 = Query(RR(node), l, r, loc);<br />}<br />return temp1 + tree[node].val;<br />}<br />int main() {<br />int n, m;<br />while (scanf("%d %d", &n, &m) != EOF && m + n) {<br />char str[2];<br />int a, b, c;<br />Build(1, 1, n);<br />while (m--) {<br />scanf("%s", str);<br />if (str[0] == 'I') {<br />//scanf("%d %d %d", &a, &b, &c);<br />readint(a);<br />readint(b);<br />readint(c);<br />Update(1, a, b, c);<br />} else {<br />//scanf("%d %d", &a, &b);<br />readint(a);<br />readint(b);<br />int loc = 0;<br />int ans = Query(1, a, b, loc);<br />printf("%d/n", ans);<br />if (loc != 0) {<br />Update(1, loc, loc, -ans);<br />}<br />}<br />}<br />}<br />}<br /> 題目沒有放出 但的確是好題值得一寫 先給出描述：                              Happy Children’s Day
Children's Day is coming. In this day, children will get a lot of candies. In MAX city, people develop an automatic candy management (ACM) system. ACM can manage N piles of candies. The system can carry out two operations.
1). I a b c (1<=a<=b<=N, 0<c<=100), ACM will add each pile numbered from a to b with c candies.
2). C a b (1<=a<=b<=N), ACM will choose the pile (from a to b) with the most candies, and give all the candies in that pile to one child. If two or more piles have the most candies, then the pile with smaller id will be chosen.
Given a series of instructions, for each operation C a b, please help the people to find out the number of candies a child can get.
Input:
Input may contain several test data sets.
For each data set, the first line contains two integers N, M(0<N,M<=10^5) where N indicates the number of piles, and M indicates the number of operations that follows.
Each of the next M lines describes one operation.
Input will be ended by N=0,M=0,which should not be processed.
NOTE: At the beginning of each data set, all of the N piles have 0 candies.
Output:
For each operation C a b, print the number of candies the child can get in a line.
Sample Input:
5 4
I 1 5 1
C 2 3
I 2 2 4
C 2 3
0 0
Sample Output:
1
4