hdu 1166 敌兵布阵（线段树）

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1166

思路：

线段树入门提，维护单点更新。

建树感觉都都差不多，主要在更新和询问两个步骤上变化挺大了，感觉还没有领悟，唉。

至于模板，没事的时候手搓几遍，慢慢就熟练了。

#include <limits.h>#include <math.h>#include <stdio.h>#include <stdlib.h>#include <string.h>#include <time.h>#include <algorithm>#include <iostream>#include <iterator>#include <queue>#include <stack>#include <string>#include <vector>#include <set>//#define ONLINE_JUDGE#define eps 1e-8#define INF 0x7fffffff                                          //INT_MAX#define inf 0x3f3f3f3f                                          //int??????????????????#define FOR(i,a) for((i)=0;i<(a);(i)++)                          //[i,a);#define MEM(a) (memset((a),0,sizeof(a)))#define sfs(a) scanf("%s",a)#define sf(a) scanf("%d",&a)#define sfI(a) scanf("%I64d",&a)#define pf(a) printf("%d\n",a)#define pfI(a) printf("%I64d\n",a)#define pfs(a) printf("%s\n",a)#define sfd(a,b) scanf("%d%d",&a,&b)#define sft(a,b,c)scanf("%d%d%d",&a,&b,&c)#define for1(i,a,b) for(int i=(a);i<b;i++)#define for2(i,a,b) for(int i=(a);i<=b;i++)#define for3(i,a,b)for(int i=(b);i>=a;i--)#define MEM1(a) memset(a,0,sizeof(a))#define MEM2(a) memset(a,-1,sizeof(a))#define MEM3(a) memset(a,0x3f,sizeof(a))#define LL __int64const double PI = acos(-1.0);template<class T> T gcd(T a, T b) { return b ? gcd(b, a % b) : a; }template<class T> T lcm(T a, T b) { return a / gcd(a, b) * b; }template<class T> inline T Min(T a, T b) { return a < b ? a : b; }template<class T> inline T Max(T a, T b) { return a > b ? a : b; }using namespace std;template<class T>T Mint(T a, T b, T c) {    if (a>b) {        if (c>b)            return b;        return c;    }    if (c>a)        return a;    return c;}template<class T>T Maxt(T a, T b, T c) {    if (a>b) {        if (c>a)            return c;        return a;    }    else if (c > b)        return c;    return b;}const int maxn=50005;int T,n,m,k;char s[10];struct node{    int l,r;    int v;}node[3*maxn+10];void build(int left,int right,int id){    node[id].l=left;    node[id].r=right;    node[id].v=0;    if(left==right)        return;    int mid=(left+right)>>1;    build(left,mid,id<<1);    build(mid+1,right,id<<1|1);}void Update(int id,int pos,int add){    int l=node[id].l,r=node[id].r;    if(l==r){        node[id].v+=add;        return ;    }    int mid=(l+r)>>1;    if(pos<=mid){        Update(id<<1,pos,add);    }    else{        Update(id<<1|1,pos,add);    }    node[id].v=node[id<<1].v+node[id<<1|1].v;}int Query(int left,int right,int id){    int l=node[id].l,r=node[id].r;    if(l==left&&r==right){        return node[id].v;    }    int mid=(l+r)>>1;    if(mid<left){        return Query(left,right,id<<1|1);    }    else if(mid>=right){        return Query(left,right,id<<1);    }    else{        int r1=Query(left,mid,id<<1);        int r2=Query(mid+1,right,id<<1|1);        return r1+r2;    }}int main() {#ifndef ONLINE_JUDGE    freopen("test.in","r",stdin);    freopen("test.out","w",stdout);#endif    int k=1;    sf(T);    while(T--){        printf("Case %d:\n",k++);        sf(n);        build(1,n,1);        for2(i,1,n){            sf(m);            Update(1,i,m);        }        int p,q;        while(sfs(s)){            if(s[0]=='E')    break;            sfd(p,q);            if(s[0]=='Q')    pf(Query(p,q,1));            else if(s[0]=='A')    Update(1,p,q);            else if(s[0]=='S')    Update(1,p,-q);        }    }    return 0;}

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