poj 3468 A Simple Problem with Integers(线段树)
http://poj.org/problem?id=3468
A Simple Problem with Integers
| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 46488 | Accepted: 13633 | |
| Case Time Limit: 2000MS | ||
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 a b c" means adding c to each of Aa, Aa+1, ... , Ab. -10000 ≤ c ≤ 10000. "Q a b" 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
【题解】：
本来以为简单的线段树，做了好久没做出来，思维还是不够清晰，然后重新理了一下思路，终于大彻大悟，嘿嘿！！！
主要是对结尾区间的传递，以及线段区间的标记，想好这两部分就不难了。。。
【code】：
1 /** 2 result:Accepted memory:8368K 3 time:1750MS language:C++ 4 code lenght: 2655B Acmer:cj 5 6 */ 7 8 #include<iostream> 9 #include<stdio.h> 10 #include<string.h> 11 #include<algorithm> 12 13 #define N 100010 14 #define lson p<<1 15 #define rson p<<1|1 16 using namespace std; 17 18 struct Nod 19 { 20 int l,r; 21 __int64 sum,val; //区间和 以及 val记录 flag为1 时 val才有值，向下更新时用 22 int flag; //标记是否为结尾区间 23 }node[N<<2]; 24 25 void building(int l,int r,int p) 26 { 27 node[p].l = l; 28 node[p].r = r; 29 node[p].flag = 0; 30 node[p].val = 0; 31 if(l==r) 32 { 33 scanf("%I64d",&node[p].sum); 34 return; 35 } 36 int mid = (l+r)>>1; 37 building(l,mid,lson); 38 building(mid+1,r,rson); 39 node[p].sum = node[lson].sum + node[rson].sum; 40 } 41 42 void update(int l,int r,int p,__int64 c) 43 { 44 node[p].sum+=(r-l+1)*c; //从上往下计算sum 45 if(node[p].l==l&&r==node[p].r) 46 { 47 if(node[p].flag) node[p].val+=c; //如果已经被标记结尾区间 val+=c 尾部区间val再增加C 48 else 49 { 50 node[p].flag = 1; //如果是新增加的结尾区间 51 node[p].val = c; //赋值区间的val 52 } 53 return; 54 } 55 if(node[p].flag) //如果遇到结尾区间，将结尾区间往下传递 56 { 57 node[p].flag = 0; 58 node[lson].val+=node[p].val; 59 node[rson].val+=node[p].val; 60 node[lson].flag = node[rson].flag = 1; //向下传递结尾区间标记 61 node[lson].sum += node[p].val*(node[lson].r-node[lson].l+1); //向下传递结尾区间sum值 62 node[rson].sum += node[p].val*(node[rson].r-node[rson].l+1); //同上 63 node[p].val = 0; 64 } 65 int mid = (node[p].l+node[p].r)>>1; 66 if(r<=mid) update(l,r,lson,c); 67 else if(l>mid) update(l,r,rson,c); 68 else 69 { 70 update(l,mid,lson,c); 71 update(mid+1,r,rson,c); 72 } 73 } 74 75 __int64 query(int l,int r,int p) 76 { 77 if(node[p].l==l&&node[p].r==r) //找到区间返回sum值 78 { 79 return node[p].sum; 80 } 81 if(node[p].flag) //如果遇到结尾区间，将结尾区间往下传递 82 { 83 node[p].flag = 0; 84 node[lson].val+=node[p].val; 85 node[rson].val+=node[p].val; 86 node[lson].flag = node[rson].flag = 1; //向下传递结尾区间标记 87 node[lson].sum += node[p].val*(node[lson].r-node[lson].l+1); //向下传递结尾区间sum值 88 node[rson].sum += node[p].val*(node[rson].r-node[rson].l+1); //同上 89 node[p].val = 0; 90 } 91 int mid = (node[p].l+node[p].r)>>1; 92 if(r<=mid) return query(l,r,lson); 93 else if(l>mid) return query(l,r,rson); 94 else return query(l,mid,lson)+query(mid+1,r,rson); 95 } 96 97 int main() 98 { 99 int n,q; 100 scanf("%d%d",&n,&q); 101 building(1,n,1); 102 char op[5]; 103 while(q--) 104 { 105 scanf("%s",op); 106 int a,b; 107 __int64 c; 108 if(op[0]=='Q') 109 { 110 scanf("%d%d",&a,&b); 111 printf("%I64d\n",query(a,b,1)); 112 } 113 else if(op[0]=='C') 114 { 115 scanf("%d%d%I64d",&a,&b,&c); 116 update(a,b,1,c); 117 } 118 } 119 return 0; 120 }
