线段树应用

(1)：区间最值查询问题 

查询区间最值下标---min

#include<iostream>    
  
using namespace std;    
    
#define MAXN 100    
#define MAXIND 256 //线段树节点个数    
    
//构建线段树,目的:得到M数组.    
void build(int node, int b, int e, int M[], int A[])    
{    
    if (b == e)    
        M[node] = b; //只有一个元素,只有一个下标    
    else    
    {     
        build(2 * node, b, (b + e) / 2, M, A);    
        build(2 * node + 1, (b + e) / 2 + 1, e, M, A);    
  
        if (A[M[2 * node]] <= A[M[2 * node + 1]])    
            M[node] = M[2 * node];    
        else    
            M[node] = M[2 * node + 1];    
    }    
}    
    
//找出区间 [i, j] 上的最小值的索引    
int query(int node, int b, int e, int M[], int A[], int i, int j)    
{    
    int p1, p2;    
    
    //查询区间和要求的区间没有交集    
    if (i > e || j < b)    
        return -1;    
  
    if (b >= i && e <= j)    
        return M[node];    
   
    p1 = query(2 * node, b, (b + e) / 2, M, A, i, j);    
    p2 = query(2 * node + 1, (b + e) / 2 + 1, e, M, A, i, j);    
    
    //return the position where the overall    
    //minimum is    
    if (p1 == -1)    
        return M[node] = p2;    
    if (p2 == -1)    
        return M[node] = p1;    
    if (A[p1] <= A[p2])    
        return M[node] = p1;    
    return M[node] = p2;    
    
}    
    
    
int main()    
{    
    int M[MAXIND]; //下标1起才有意义,否则不是二叉树,保存下标编号节点对应区间最小值的下标.    
    memset(M,-1,sizeof(M));    
    int a[]={3,4,5,7,2,1,0,3,4,5};    
    build(1, 0, sizeof(a)/sizeof(a[0])-1, M, a);    
    cout<<query(1, 0, sizeof(a)/sizeof(a[0])-1, M, a, 0, 5)<<endl;    
    return 0;    
}   

(2)：连续区间修改或者单节点更新的动态查询问题 

#include <cstdio>    
#include <algorithm>    
using namespace std;    
     
#define lson l , m , rt << 1    
#define rson m + 1 , r , rt << 1 | 1   
#define root 1 , N , 1   
#define LL long long    
const int maxn = 111111;    
LL add[maxn<<2];    
LL sum[maxn<<2];    
void PushUp(int rt) {    
    sum[rt] = sum[rt<<1] + sum[rt<<1|1];    
}    
void PushDown(int rt,int m) {    
    if (add[rt]) {    
        add[rt<<1] += add[rt];    
        add[rt<<1|1] += add[rt];    
        sum[rt<<1] += add[rt] * (m - (m >> 1));    
        sum[rt<<1|1] += add[rt] * (m >> 1);    
        add[rt] = 0;    
    }    
}    
void build(int l,int r,int rt) {    
    add[rt] = 0;    
    if (l == r) {    
        scanf("%lld",&sum[rt]);    
        return ;    
    }    
    int m = (l + r) >> 1;    
    build(lson);    
    build(rson);    
    PushUp(rt);    
}    
void update(int L,int R,int c,int l,int r,int rt) {    
    if (L <= l && r <= R) {    
        add[rt] += c;    
        sum[rt] += (LL)c * (r - l + 1);    
        return ;    
    }    
    PushDown(rt , r - l + 1);    
    int m = (l + r) >> 1;    
    if (L <= m) update(L , R , c , lson);    
    if (m < R) update(L , R , c , rson);    
    PushUp(rt);    
}    
LL query(int L,int R,int l,int r,int rt) {    
    if (L <= l && r <= R) {    
        return sum[rt];    
    }    
    PushDown(rt , r - l + 1);    
    int m = (l + r) >> 1;    
    LL ret = 0;    
    if (L <= m) ret += query(L , R , lson);    
    if (m < R) ret += query(L , R , rson);    
    return ret;    
}    
int main() {    
    int N , Q;    
    scanf("%d%d",&N,&Q);    
    build(root);    
    while (Q --) {    
        char op[2];    
        int a , b , c;    
        scanf("%s",op);    
        if (op[0] == 'Q') {    
            scanf("%d%d",&a,&b);    
            printf("%lld
",query(a , b ,root));    
        } else {    
            scanf("%d%d%d",&a,&b,&c);    
            update(a , b , c , root);    
        }    
    }    
    return 0;    
}    

摘自：http://blog.csdn.net/metalseed/article/details/8039326