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hdu 1166 敌兵布阵 区间和

2016-12-13 16:27 260 查看
Problem:

给n个节点,有很多命令,两种命令,一种是更新节点的值,一种是区间求和:

Solution:

利用线段树堆区间求和

note:

要关闭cin的同步,不然会超时。

#include<cstdio>
#include<iostream>
#include<sstream>
#include<cstdlib>
#include<cmath>
#include<cctype>
#include<string>
#include<cstring>
#include<algorithm>
#include<stack>
#include<queue>
#include<set>
#include<map>
#include<ctime>
#include<vector>
#include<fstream>
#include<list>
using namespace std;

#define ms(s) memset(s,0,sizeof(s))
typedef unsigned long long ULL;
typedef long long LL;

const double PI = 3.141592653589;
const int INF = 0x3fffffff;

#define maxn 50000

struct node{
int l, r, sum;
int mid(){
return (l+r)/2;
}
};

node Tree[maxn*4];
int value[maxn+10];
int flag;

//初始化树,根节点是1
void init_tree(int root, int l, int r){
Tree[root].l = l;
Tree[root].r = r;
if(l == r)
Tree[root].sum = value[l];
else{
init_tree(2*root, l, (l+r)/2);
init_tree(2*root+1, (l+r)/2 + 1, r);

Tree[root].sum = Tree[2*root].sum + Tree[2*root+1].sum;
}
}

//查找和
int query_tree(int root, int l, int r){
int m = Tree[root].mid();
if(l == Tree[root].l && r == Tree[root].r)
return Tree[root].sum;
else{
if(l > m)
return query_tree(2*root+1, l, r);
else if(r <= m)
return query_tree(2*root, l, r);
else
return query_tree(2*root, l, m) + query_tree(2*root+1, m+1, r);
}
}

void update_tree(int root, int idx, int v){
if(Tree[root].l == Tree[root].r)
Tree[root].sum += flag*v;
else{
Tree[root].sum += flag*v;

if(idx <= Tree[root].mid())
update_tree(2*root, idx, v);
else
update_tree(2*root+1, idx, v);
}
}

int main(){
//        freopen("/Users/really/Documents/code/input","r",stdin);
//        freopen("/Users/really/Documents/code/output","w",stdout);
ios::sync_with_stdio(false);

int t, n;
int a, b;
string s;
int res;
cin >> t;
for(int k = 1; k <= t; ++k){
cout << "Case " << k << ":"<<endl;
cin >> n;
for(int i = 1; i <= n; ++i)
cin >> value[i];
init_tree(1, 1, n);
while(cin >> s){
if(s == "Add"){
cin >> a >> b;
flag = 1;
update_tree(1, a, b);
}
else if(s == "Sub"){
cin >> a >> b;
flag = -1;
update_tree(1, a, b);
}
else if(s == "Query"){
cin >> a >> b;
res = query_tree(1, a, b);
cout << res << endl;
}
else
break;
}
}

return 0;
}
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