ccnu-线段树-简单的区间更新（三题）

题目一：http://poj.org/problem?id=3468

Description

You have N integers, A1, A2, ... , AN. 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, ... , AN. -1000000000 ≤ Ai ≤ 1000000000.
Each of the next Q lines represents an operation.
"C abc" means adding c to each of Aa, Aa+1, ... , Ab. -10000 ≤ c ≤ 10000.
"Q ab" means querying the sum of Aa, Aa+1, ... , Ab.

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

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
Hint

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

裸的区间更新，关于lazy标记，还是老话，在下一次更新或询问时才把区间信息传递（pushdown）下去。

代码：

#include<iostream>#include<cstdio>#include<string>using namespace std;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define Max  100020long long a,b,q,n;long long sum[Max<<2],c,lazy[Max<<2];void pushup(int rt){    sum[rt] = sum[rt<<1] + sum[rt<<1|1];}void pushdown(int rt,int m){    if(lazy[rt])    {        lazy[rt<<1] += lazy[rt];        lazy[rt<<1|1] += lazy[rt];        sum[rt<<1] += lazy[rt] * (m - (m>>1));        sum[rt<<1|1] += lazy[rt] * (m>>1);        lazy[rt] = 0;    }}void build(int l,int r,int rt){    sum[rt] = 0;    if(l == r)    {        scanf("%lld",&sum[rt]);        return;    }    int m = (r + l)>>1;    build(lson);    build(rson);    pushup(rt);}void Add(int L,int R,int v,int l,int r,int rt){    if(L <= l&&r <= R)    {        lazy[rt] += v;        sum[rt] += (long long)(r - l + 1) * v;        return ;    }    pushdown(rt,r-l+1);    int m = (r + l)>>1;    if(L <= m) Add(L,R,v,lson);            if(m < R) Add(L,R,v,rson);    pushup(rt);}long long query(int L,int R,int l,int r,int rt){    if(L <= l&&r <= R)    {        return sum[rt];    }    long long rec = 0;    int m = (r + l)>>1;    pushdown(rt,r-l+1);    if(L <= m) rec += query(L,R,lson);    if(m < R) rec += query(L,R,rson);    return rec;}int main(){    scanf("%d%d",&n,&q);    build(1,n,1);    while(q--)    {        char qus[2];        scanf("%s",&qus);        if(qus[0] == 'C')        {            scanf("%d%d%d",&a,&b,&c);            Add(a,b,c,1,n,1);        }        else        {            scanf("%d%d",&a,&b);            printf("%lld\n",query(a,b,1,n,1));        }    }    return 0;}

题型二：http://acm.hdu.edu.cn/showproblem.php?pid=1698

把上题的求和改成更新区间值就可以了。

代码如下：

#include<iostream>#include<cstdio>#define Max 100010#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1using namespace std;int sum[Max<<2],lazy[Max<<2];void pushup(int rt){    sum[rt] = sum[rt<<1] + sum[rt<<1|1];}void pushdown(int rt,int INL){    if(lazy[rt])    {        lazy[rt<<1] = lazy[rt<<1|1] = lazy[rt];        sum[rt<<1] = lazy[rt] * (INL - (INL>>1));        sum[rt<<1|1] = lazy[rt] * (INL>>1);        lazy[rt] = 0;    }}void build(int l,int r,int rt){    lazy[rt] = 0;    sum[rt] = 1;    if(r == l)    {        return;    }    int m = (r + l) >>  1;    build(lson);    build(rson);    pushup(rt);}void update(int L,int R,int val,int l,int r,int rt){    if(L <= l&&r <= R)    {        sum[rt] = val * (r-l+1);        lazy[rt] = val;        return;    }    pushdown(rt,r-l+1);    int m = (r + l)>>1;    if(L <= m) update(L,R,val,lson);    if(m < R) update(L,R,val,rson);    pushup(rt);}int main(){    int T,n,x,y,z,q;    scanf("%d",&T);    int lala = T;    while(T--)    {        scanf("%d",&n);        build(1,n,1);        scanf("%d",&q);        while(q--)        {            scanf("%d%d%d",&x,&y,&z);            update(x,y,z,1,n,1);        }        printf("Case %d: The total value of the hook is %d.\n",(lala-T),sum[1]);    }    return 0;}

【题型三-hdu-ColorTheBall】http://acm.hdu.edu.cn/showproblem.php?pid=1556

题意：把a,b区间的 气球都涂上颜色，输出每个气球被涂色的次数。

思路：每次更新只记录到所更新的区间便不再往下记录更新，不需要pushup，只需要在输出时pushdown更新一次到每个节点就可以了

经验教训： 不要用cout， 不要用cout， 不要用cout！

#include<iostream>#include<cstdio>#include<algorithm>using namespace std;#define maxn 100010#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1int lazy[maxn<<4]; int n;//int index[maxn];void build(int l, int r, int rt){  int m;  m = (l + r)>>1;  lazy[rt] = 0;  if(l == r) return;  build(lson);  build(rson);}void pushdown(int rt){  if(lazy[rt])  {    lazy[rt<<1] += lazy[rt];    lazy[rt<<1|1] += lazy[rt];    lazy[rt] = 0;  }}void paint(int l, int r, int rt, int L, int R){  if(L <= l && r <= R)  {    lazy[rt]++;    return ;  }  int m = (r+l)>>1;  if(L <= m) paint(lson, L,R);   if(m < R) paint(rson, L, R);}void finalQuery(int l, int r, int rt){  int m = (l+r)>>1;  if(l == r)  {     printf("%d%c",lazy[rt],l==n?'\n':' ');     return;  }  pushdown(rt);   //在查询的时候才pushdown  finalQuery(lson);  finalQuery(rson);}int main(){ while(scanf("%d", &n) != EOF&&n) {   build(1,n,1);   for(int i = 0; i < n; i++)   {     int a, b;     scanf("%d%d", &a, &b);     paint(1, n, 1, a, b);     //cout<<"-"<<endl;   }   finalQuery(1, n, 1); } return 0;}