poj 3468 A Simple Problem with Integers 线段树 题解《挑战程序设计竞赛》

地址 http://poj.org/problem?id=3468 

线段树模板 

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线段树

#include <iostream>
#include <vector>
#include <math.h>
#include <algorithm>


using namespace std;

/*
Sample Input
10 5
1 2 3 4 5 6 7 8 9 10
Q 4 4
Q 1 10
Q 2 4
C 3 6 3
Q 2 4
Sample Output
4
55
9
15
*/
typedef long long ll;

const int  DAT_SIZE = (1 << 18) - 1;

const int MAX_N = 100000010;

//输入
int N, Q;
int A[MAX_N];
char T[MAX_N];
int L[MAX_N], R[MAX_N], X[MAX_N];

//线段树
ll dat_a[DAT_SIZE], dat_b[DAT_SIZE];

//对区间[a,b]同时加x
//k是节点的编号 对应的空间是[l,r)
void add(int a, int b, int x, int k, int l, int r)
{
    if (a <= l && r <= b) {
        dat_a[k] += x;
    }
    else if (l < b && a < r) {
        dat_b[k] += (min(b,r)-max(a,l))* x;
        add(a, b, x, k * 2 + 1, l, (l + r) / 2);
        add(a, b, x, k * 2 + 2, (l + r) / 2, r);
    }
}

//计算[a,b)的和
//k是节点的编号 对应的区间是[l,r)
ll sum(int a, int b, int k, int l, int r)
{
    if (b <= l || r <= a) {
        return 0;
    }
    else if(a <= l && r <= b){
        return dat_a[k] * (r - l) + dat_b[k];
    }
    else {
        ll res = (min(b,r)-max(a,l)) * dat_a[k];
        res += sum(a, b, k * 2 + 1, l, (l + r) / 2);
        res += sum(a, b, k * 2 + 2, (l + r) / 2, r);
        return res;
    }
}

void solve()
{
    for (int i = 0; i < N; i++) {
        add(i, i + 1, A[i], 0, 0, N);
    }
    for (int i = 0; i < Q; i++) {
        if (T[i] == 'C') {
            add(L[i]-1, R[i] , X[i], 0, 0, N);
        }
        else {
            printf("%lld
",sum(L[i]-1,R[i],0,0,N));
        }
    }
}



int main()
{
    cin >> N >> Q;

    for (int i = 0; i < N; i++) {
        cin >> A[i];
    }

    for (int i = 0; i < Q; i++) {
        cin >> T[i];
        if (T[i] == 'C')
            cin >> L[i] >> R[i] >> X[i];
        else
            cin >> L[i] >> R[i];
    }

    solve();

    return 0;
}

#include <iostream>
#include <vector>


using namespace std;

/*
Sample Input
10 5
1 2 3 4 5 6 7 8 9 10
Q 4 4
Q 1 10
Q 2 4
C 3 6 3
Q 2 4
Sample Output
4
55
9
15
*/

typedef long long ll;

int n, m;

const int MAX_N  = 1000010;
int input[MAX_N];

struct Node {
    int l, r;
    ll data;
};

struct Node stree[MAX_N * 4];


void build(int idx, int l, int r)
{
    stree[idx].l = l;
    stree[idx].r = r;
    if (l == r) {
        //叶子节点
        stree[idx].data = input[l];
        return;
    }

    int mid = (l + r) / 2;
    build(idx * 2, l,mid);
    build(idx * 2 + 1, mid + 1, r);

    stree[idx].data = stree[idx * 2].data + stree[idx * 2 + 1].data;
}

ll query(int idx, int start, int end, int l, int r)
{
    if (l == stree[idx].l && r == stree[idx].r) {
        return stree[idx].data;
    }

    int mid = (start + end) / 2;
    if (l <= mid && r <= mid) {
        return query(idx * 2, start, mid, l, r);
    }
    else if (l > mid && r > mid) {
        return query(idx * 2 + 1, mid + 1, end, l, r);
    }
    else {
        ll res = 0;
        res += query(idx * 2 , start, mid, l, mid);
        res += query(idx * 2 + 1, mid + 1, end, mid + 1, r);
        return res;
    }

}

void add(int idx, int l, int r, int v)
{
    if (l == r && stree[idx].l == stree[idx].r && stree[idx].l == l) {
        //叶子节点
        stree[idx].data += v;
        return;
    }

    int mid = (stree[idx].l + stree[idx].r) / 2;
    if (l <= mid && r <= mid) {
        add(idx * 2, l, r,v);
    }
    else if (l > mid && r > mid) {
        add(idx * 2 + 1, l, r,v);
    }
    else {
        add(idx * 2 , l, mid, v);
        add(idx * 2 + 1, mid+1, r, v);
    }

    stree[idx].data = stree[idx*2].data + stree[idx*2+1].data;
}



int main()
{
    int n, m;
    cin >> n >> m;
    for (int i = 1; i <= n; i++) {
        cin >> input[i];
    }

    build(1, 1, n);

    for (int i = 0; i < m; i++) {
        char t;
        int l, r, v;
        cin >> t;
        if (t == 'C') {
            cin >> l >> r >> v;
            add(1, l, r, v);
        }
        else if (t == 'Q') {
            cin >> l >> r;
            cout << query(1, 1, n, l, r) << endl;
        }
    }


    return 0;
}