kuangbin線段樹專題解析

賽前再鞏固一下線段樹，又做了一遍kuangbin線段樹專題

專題鏈接：kuangbin專題題目一覽

1. HDU 1166 敵兵佈陣

區間維護和，單點修改

#include <stdio.h>
using namespace std;
const int MAXN=1e5 + 10;
int origin[MAXN], tree[MAXN<<2];

void pushup(int p) {
    tree[p] = tree[p << 1] + tree[p << 1 | 1];
}

void build(int p, int l, int r) {
    if (l == r) {
        tree[p] = origin[l];
        return;
    }
    int mid = (l + r) >> 1;
    build(p << 1, l, mid);
    build(p << 1 | 1, mid + 1, r);
    pushup(p);
}

void update_node(int p, int l,int r, int q,int v) {
    if (l == r) {   //查詢到點
        tree[p] += v;
        return;
    }
    int mid = (l + r) >> 1;
    if (q > mid) update_node(p << 1 | 1, mid + 1, r, q, v);
    else update_node(p << 1, l, mid, q, v);
    pushup(p);
}

int query(int p, int l, int r, int ql, int qr) {
    if (ql <= l && r <= qr)  return tree[p];    //被包含在詢問區域內的區間(有效的部分)
    int mid = (l + r) >> 1;
    int temp = 0;
    if (qr > mid) temp += query(p << 1 | 1, mid + 1, r, ql, qr);    //分塊切割出有效的部分(已忽略無效部分)
    if (ql <= mid) temp += query(p << 1, l, mid, ql, qr);
    return temp;
}

int main () {
    int T;
    scanf("%d", &T);
    for (int k = 1; k <= T; k++) {
        int n, i, j;
        char s[10];
        scanf("%d", &n);
        for (int i = 1; i <= n; i++)
            scanf("%d", origin + i);
        build(1, 1, n);
        printf("Case %d:\n", k);
        while(scanf("%s", s) && s[0] != 'E') {
            scanf("%d%d", &i, &j);
            if (s[0] == 'A')
                update_node(1, 1, n, i, j);
            else if (s[0] == 'S')
                update_node(1, 1, n, i, -j);
            else
                printf("%d\n", query(1, 1, n, i, j));
        }
    }
}

2. HDU 1754 I Hate It

維護區間最大值，單點修改

#include <stdio.h>
#include <algorithm>
using namespace std;
const int MAXN = 2e5 + 10;


int origin[MAXN], tree[MAXN<<2], lazy[MAXN<<2];

void pushup(int p) {
    tree[p] = max(tree[p << 1], tree[p << 1 | 1]);
}

void build(int p, int l, int r) {
    if (l == r) {
        tree[p] = origin[l];
        return;
    }
    int mid = (l + r) >> 1;
    build(p << 1, l, mid);
    build(p << 1 | 1, mid + 1, r);
    pushup(p);
}

void update_node(int p, int l,int r, int q,int v) {
    if (l == r) {   //查詢到點
        tree[p] = v;
        return;
    }
    int mid = (l + r) >> 1;
    if (q > mid) update_node(p << 1 | 1, mid + 1, r, q, v);
    else update_node(p << 1, l, mid, q, v);
    pushup(p);
}

int query(int p, int l, int r, int ql, int qr) {
    if (ql <= l && r <= qr)  return tree[p];    //被包含在詢問區域內的區間(有效的部分)
    int mid = (l + r) >> 1;
    int temp = -1;
    if (qr > mid) temp = max(temp, query(p << 1 | 1, mid + 1, r, ql, qr));    //分塊切割出有效的部分(已忽略無效部分)
    if (ql <= mid) temp = max(temp, query(p << 1, l, mid, ql, qr));
    return temp;
}

int main() {
    int n, m;
    while(scanf("%d%d", &n, &m) != EOF) {
    for (int i = 1; i <= n; i++)
        scanf("%d", origin + i);
    build(1, 1, n);
    while(m--) {
        char c[3];
        int i, j;
        scanf("%s%d%d", c, &i, &j);
        if (c[0] == 'Q')
            printf("%d\n", query(1, 1, n, i, j));
        else
            update_node(1, 1, n, i, j);
    }
    }
}

3. POJ 3468 A Simple Problem with Integers

維護區間和，區間修改

#include <stdio.h>
using namespace std;
typedef long long ll;
const int MAXN=1e5 + 10;
ll origin[MAXN], tree[MAXN<<2], lazy[MAXN<<2];

void pushup(int p) {
    tree[p] = tree[p << 1] + tree[p << 1 | 1];
}

void pushdown(int p, int l, int r) {
    if (lazy[p]) {
        lazy[p << 1] += lazy[p];
        lazy[p << 1 | 1] += lazy[p];
        int mid = (l + r) >> 1;
        tree[p << 1] += lazy[p] * (mid - l + 1);  //左右兒子結點均按照需要加的值總和更新結點信息
        tree[p << 1 | 1] += lazy[p] * (r - mid);
        lazy[p] = 0;
    }
}

void build(int p, int l, int r) {
    if (l == r) {
        tree[p] = origin[l];
        return;
    }
    int mid = (l + r) >> 1;
    build(p << 1, l, mid);
    build(p << 1 | 1, mid + 1, r);
    pushup(p);
}

void update(int p, int l, int r, int ql, int qr, ll v) {   //區間更新
    if (ql <= l && r <= qr) {       //當前區間在更新區間內
        tree[p] += v * (r - l + 1);
        lazy[p] += v;
     //   pushdown(p, l, r);
        return;
    }
    int mid = (l + r) >> 1;
    pushdown(p, l, r);
    if (ql <= mid)  update(p << 1, l, mid, ql, qr, v);
    if (qr > mid)  update(p << 1 | 1, mid + 1, r, ql, qr, v);
    pushup(p);
}


ll query(int p, int l, int r, int ql, int qr) {
    if (ql <= l && r <= qr)  return tree[p];    //被包含在詢問區域內的區間(有效的部分)
    int mid = (l + r) >> 1;
    pushdown(p, l, r);        //有區間更新時才需要
    ll temp = 0;
    if (qr > mid) temp += query(p << 1 | 1, mid + 1, r, ql, qr);    //分塊切割出有效的部分(已忽略無效部分)
    if (ql <= mid) temp += query(p << 1, l, mid, ql, qr);
    return temp;
}

int main () {

    int n, m;
    while(scanf("%d%d", &n, &m)!= EOF) {
    int i, j;
        ll c;
        char s[5];
        for (int i = 1; i <= n; i++)
            scanf("%lld", origin + i);
        build(1, 1, n);
        while(m--) {
            scanf("%s", s);
            if (s[0] == 'Q') {
                scanf("%d%d", &i, &j);
                printf("%lld\n", query(1, 1, n, i, j));
            }
            else {
                scanf("%d%d%lld", &i, &j, &c);
                update(1, 1, n, i, j, c);
            }
        }
    }
}

4. POJ 2528 Mayor's posters

見題解：POJ 2528 Mayor's posters [ 離散化 + 線段樹 ]

5. HDU 1698 Just a Hook

維護區間和，區間修改

#include <stdio.h>
#include <memory.h>
using namespace std;
const int MAXN=1e5 + 10;
int origin[MAXN], tree[MAXN<<2], lazy[MAXN<<2];

void pushup(int p) {
    tree[p] = tree[p << 1] + tree[p << 1 | 1];
}

void pushdown(int p, int l, int r) {
    if (lazy[p]) {
        lazy[p << 1] = lazy[p];
        lazy[p << 1 | 1] = lazy[p];
        int mid = (l + r) >> 1;
        tree[p << 1] = lazy[p] * (mid - l + 1);  //左右兒子結點均按照需要加的值總和更新結點信息
        tree[p << 1 | 1] = lazy[p] * (r - mid);
        lazy[p] = 0;
    }
}

void build(int p, int l, int r) {
    if (l == r) {
        tree[p] = origin[l];
        return;
    }
    int mid = (l + r) >> 1;
    build(p << 1, l, mid);
    build(p << 1 | 1, mid + 1, r);
    pushup(p);
}

void update(int p, int l, int r, int ql, int qr, int v) {   //區間更新
    if (ql <= l && r <= qr) {       //當前區間在更新區間內
        tree[p] = v * (r - l + 1);
        lazy[p] = v;
        return;
    }
    int mid = (l + r) >> 1;
    pushdown(p, l, r);
    if (ql <= mid)  update(p << 1, l, mid, ql, qr, v);
    if (qr > mid)  update(p << 1 | 1, mid + 1, r, ql, qr, v);
    pushup(p);
}


int query(int p, int l, int r, int ql, int qr) {
    if (ql <= l && r <= qr)  return tree[p];    //被包含在詢問區域內的區間(有效的部分)
    int mid = (l + r) >> 1;
    pushdown(p, l, r);        //有區間更新時才需要
    int temp = 0;
    if (qr > mid) temp += query(p << 1 | 1, mid + 1, r, ql, qr);    //分塊切割出有效的部分(已忽略無效部分)
    if (ql <= mid) temp += query(p << 1, l, mid, ql, qr);
    return temp;
}

int main () {

    int T;
    scanf("%d", &T);
    for (int k = 1; k <= T; k++) {
        int n, m;
    scanf("%d%d", &n, &m);
    int i, j, c;
        for (int i = 1; i <= n; i++)
           origin[i] = 1;
        memset(lazy, 0, sizeof(lazy));
        build(1, 1, n);
        while(m--) {
                scanf("%d%d%d", &i, &j, &c);
                update(1, 1, n, i, j, c);
            }
            printf("Case %d: The total value of the hook is %d.\n", k, query(1,1,n,1, n));
        }
    }

6. ZOJ 1610 Count the Colors

見題解：ZOJ 1610 Count the Colors [ 線段樹 + 區間染色 ]

7. POJ 3264 Balanced Lineup

維護區間最小值、最大值，不需要更新。用RMQ做更快

#include <stdio.h>
#include <algorithm>
using namespace std;
const int MAXN = 5e4 + 10;
const int INF = 0x3f3f3f3f;

int origin[MAXN];
struct node {
    int max;
    int min;
}tree[MAXN<<2];

void pushup(int p) {
    tree[p].max = max(tree[p << 1].max, tree[p << 1 | 1].max);
    tree[p].min = min(tree[p << 1].min, tree[p << 1 | 1].min);
}

void build(int p, int l, int r) {
    if (l == r) {
        tree[p].max = tree[p].min = origin[l];
        return;
    }
    int mid = (l + r) >> 1;
    build(p << 1, l, mid);
    build(p << 1 | 1, mid + 1, r);
    pushup(p);
}

node query(int p, int l, int r, int ql, int qr) {
    if (ql <= l && r <= qr)  return tree[p];    //被包含在詢問區域內的區間(有效的部分)
    int mid = (l + r) >> 1;
    node temp, temp2;
    temp.max = -INF;
    temp.min = INF;
    if (qr > mid) {
        //一定要注意：這裏應該先存下temp2,而不是每次都query出來，可以省很多時間，否則會TLE
        temp2 = query(p << 1 | 1, mid + 1, r, ql, qr);
        temp.max = max(temp.max, temp2.max);
        temp.min = min(temp.min, temp2.min);
    }
    if (ql <= mid) {
        temp2 = query(p << 1, l, mid, ql, qr);
        temp.max = max(temp.max, temp2.max);
        temp.min = min(temp.min, temp2.min);
    }
    return temp;
}

int main () {
    int n, q;
    scanf("%d%d", &n, &q);
    for (int i = 1; i <= n; i++)   //origin數組的下標從1開始
        scanf("%d", origin + i);
    build(1, 1, n);     //拿區間 1, n 的數據建樹
    while (q--) {
        int i, j;
        scanf("%d%d", &i, &j);
        node temp = query(1, 1, n, i, j);
        printf("%d\n", temp.max - temp.min);
    }
}