POJ 1890 - A Simple Problem with Integers （线段树 Splay）

A Simple Problem with Integers

Time Limit: 5000MS Memory Limit: 131072KTotal Submissions: 64159 Accepted: 19757Case Time Limit: 2000MS

Description

You have N integers, A₁, A₂, ... , A_N. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval.

Input

The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000.
The second line contains N numbers, the initial values of A₁, A₂, ... , A_N. -1000000000 ≤ A_i ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of A_a, A_a+1, ... , A_b. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of A_a, A_a+1, ... , A_b.

Output

You need to answer all Q commands in order. One answer in a line.

Sample Input

10 51 2 3 4 5 6 7 8 9 10Q 4 4Q 1 10Q 2 4C 3 6 3Q 2 4

Sample Output

455915

Hint

The sums may exceed the range of 32-bit integers.

Source

POJ Monthly--2007.11.25, Yang Yi

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题意：

n个整数m个操作。每次操作：

Q a b 询问[a, b]的和

C a b c [a, b] 所有值+ c

线段树做非常简单 这里我学习Splay 用来试一下

第一道Splay

#include <cstdio>#include <iostream>#include <vector>#include <algorithm>#include <cstring>#include <string>#include <map>#include <cmath>#include <queue>#include <set>using namespace std;//#define WIN#ifdef WINtypedef __int64 LL;#define iform "%I64d"#define oform "%I64d\n"#define oform1 "%I64d"#elsetypedef long long LL;#define iform "%lld"#define oform "%lld\n"#define oform1 "%lld"#endif#define S64I(a) scanf(iform, &(a))#define P64I(a) printf(oform, (a))#define P64I1(a) printf(oform1, (a))#define REP(i, n) for(int (i)=0; (i)<n; (i)++)#define REP1(i, n) for(int (i)=1; (i)<=(n); (i)++)#define FOR(i, s, t) for(int (i)=(s); (i)<=(t); (i)++)#define keyTree (ch[ch[root][1]][0])const int INF = 0x3f3f3f3f;const double eps = 1e-9;const double PI = (4.0*atan(1.0));const int maxn = 100000 + 20;int ch[maxn][2], val[maxn], pre[maxn], size[maxn], S[maxn];int root, top1, top2;LL sumv[maxn], addv[maxn];// debugvoid Treaval(int x) {    if(x) {        Treaval(ch[x][0]);        printf("结点%2d : 左儿子 %2d 右儿子 %2d 父结点 %2d size = %2d ,val = %2d\n",x,ch[x][0],ch[x][1],pre[x],size[x],val[x]);        Treaval(ch[x][1]);    }}void debug() {    printf("root=%d\n",root);    Treaval(root);}void New(int & r, int fa, int v) {    if(top2) r = S[--top2];    else r = ++top1;    pre[r] = fa;    size[r] = 1;    ch[r][0] = ch[r][1] = 0;    sumv[r] = val[r] = v;    addv[r] = 0;}void push_up(int r) {    size[r] = size[ch[r][0]] + size[ch[r][1]] + 1;    sumv[r] = sumv[ch[r][0]] + sumv[ch[r][1]] + val[r];}void push_down(int r) {    if(addv[r]) {        int lson = ch[r][0], rson = ch[r][1];        addv[lson] += addv[r];        val[lson] += addv[r];        addv[rson] += addv[r];        val[rson] += addv[r];        sumv[lson] += addv[r] * size[lson];        sumv[rson] += addv[r] * size[rson];        addv[r] = 0;    }}void Rotate(int x, int d) {    int y = pre[x];    push_down(x);    push_down(y);    ch[y][!d] = ch[x][d];    pre[ch[x][d]] = y;    if(pre[y]) ch[pre[y]][ch[pre[y]][1] == y] = x;    pre[x] = pre[y];    ch[x][d] = y;    pre[y] = x;    push_up(y);}void Splay(int x, int goal) {    push_down(x);    while(pre[x] != goal) {        if(pre[pre[x]] == goal)            Rotate(x, ch[pre[x]][0] == x);        else {            int y = pre[x];            int d = ch[pre[y]][0] == y;            if(ch[y][d] == x) {                Rotate(x, !d);                Rotate(x, d);            } else {                Rotate(y, d);                Rotate(x, d);            }        }    }    push_up(x);    if(goal == 0) root = x;}void RotateTo(int k, int goal) {    int r = root;    push_down(r);    while(size[ch[r][0]] != k) {        if(k < size[ch[r][0]]) r = ch[r][0];        else {            k -= size[ch[r][0]] + 1;            r = ch[r][1];        }        push_down(r);    }    Splay(r, goal);}int que[maxn];void erase(int x) {    int y = pre[x];    int head = 0, tail = 0;    for(que[tail++] = x; head < tail; head++) {        int r = que[head];        S[top2++] = r;        if(ch[r][0]) que[tail++] = ch[r][0];        if(ch[r][1]) que[tail++] = ch[r][1];    }    ch[y][ch[y][1] == x] = 0;    push_up(y);}int A[maxn];void build(int & r, int L, int R, int fa) {    if(L > R) return ;    int M = (L + R) / 2;    New(r, fa, A[M]);    if(L < M) build(ch[r][0], L, M-1, r);    if(M < R) build(ch[r][1], M+1, R, r);    push_up(r);}void init(int n) {    root = top1 = top2 = 0;    ch[0][0] = ch[0][1] = size[0] = pre[0] = 0;    addv[0] = sumv[0] = 0;    New(root, 0, -1);    New(ch[root][1], root, -1);    size[root] = 2;    build(keyTree, 1, n, ch[root][1]);    push_up(ch[root][1]);    push_up(root);}int getPre(int r) {    int t = ch[r][0];    if(!t) return INF;    while(ch[t][1]) t = ch[t][1];    return val[t];}int getNext(int r) {    int t = ch[r][1];    if(!t) return INF;    while(ch[t][0]) t = ch[t][0];    return val[t];}LL query(int L, int R) {    RotateTo(L-1, 0);    RotateTo(R+1, root);    return sumv[keyTree];}void update(int L, int R, int v) {    RotateTo(L-1, 0);    RotateTo(R+1, root);    addv[keyTree] += v;    val[keyTree] += v;    sumv[keyTree] += (LL) v * size[keyTree];}char str[10];int main() {    int n, m;    while(scanf("%d%d", &n, &m) != EOF) {        for(int i=1; i<=n; i++) scanf("%d", &A[i]);        init(n);        while(m--) {            int a, b, c;            scanf("%s%d%d", str, &a, &b);            if(str[0] == 'Q') {                LL ans = query(a, b);                P64I(ans);            } else {                scanf("%d", &c);                update(a, b, c);            }        }    }    return 0;}