hdu 1166 敌兵布阵

来源：互联网发布：php ruby perl python 编辑：程序博客网时间：2024/06/06 10:53

线段树查询和更新的一道简单题，注意右移>> 和左移<<运算的优先级 (它们的优先级比 + 、- 还低)，另外就是处理输入输出时不要使用cin、cout。

代码:

#include <iostream>#include <stdlib.h>#include <stdio.h>#include <string>#include <string.h>#include <algorithm>using namespace std;const int maxn = 50000 + 10;int tree[maxn * 4];//线段树,除过叶节点外，其余为所对应区间总人数int casenum;int N;//N个工兵阵地int peo[maxn];//工兵阵地人数char str[30];void build_tree(int node,int lp,int rp)//结点node 所对应区间[lp,rp]{if (lp == rp)tree[node] = peo[lp];else{int mid = (lp + rp) >> 1;build_tree(node << 1, lp, mid);build_tree((node << 1) + 1, mid + 1, rp);tree[node] = tree[node << 1] + tree[(node << 1) + 1];}}//向结点 node 区间[lp,rp]增加army val，void add_peo(int node, int lp, int rp,int army,int val){//army 在区间[lp,rp]之外if (army < lp || army > rp)return;else if (lp == rp && lp == army){tree[node] += val;return;}tree[node] += val;int mid = (lp + rp) >> 1;add_peo(node << 1, lp, mid, army, val);add_peo((node << 1) + 1, mid + 1, rp, army, val);}////void sub_peo(int node, int lp, int rp, int army, int val)//{//if (army < lp || army > rp)//return;//else if (lp == rp && lp == army)//{//tree[node] -= val;//return;//}////tree[node] -= val;//int mid = (lp + rp) / 2;//sub_peo(node * 2, lp, mid, army, val);//sub_peo(node * 2 + 1, mid + 1, rp, army, val);//}int query(int node, int tl, int tr, int lp, int rp){//如果 【lp，rp】 与区间[tl,tr]无任何交集if (tr < lp || tl > rp)return 0;//如果 [lp,rp]在区间 [tl,tr]范围内if (tl <= lp && rp <= tr)return tree[node];int mid = (lp + rp) >> 1;return query(node << 1, tl, tr, lp, mid) +query((node <<1) + 1, tl, tr, mid + 1, rp);}int main(){freopen("in.txt", "r", stdin);freopen("out.txt", "w", stdout);int m, n,index = 1;//指令后面的数字//cin >> casenum;scanf("%d", &casenum);while (casenum--){//cin >> N;scanf("%d", &N);for (int i = 1; i <= N; i++){//cin >> peo[i];scanf("%d",&peo[i]);}//cout << "Case " << index++ << ":" << endl;printf("Case %d:\n", index++);build_tree(1, 1, N);//cin >> str;scanf("%s", str);while ('E' != str[0]){scanf("%d %d", &m, &n);if ('Q' == str[0]){printf("%d\n", query(1, m, n, 1, N));}else if ('A' == str[0])add_peo(1, 1, N, m, n);elseadd_peo(1, 1, N, m, -n);scanf("%s", str);}//while end}//while casenumreturn 0;}

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