ACM——A Simple Problem with Integers（线段树的精华版）

  1 #include<cstdio>
  2 #include<cstring>
  3 #include<iostream>
  4 
  5 using namespace std;
  6 
  7 const int MAXN=500000;
  8 
  9 struct student
 10 {
 11     int left,right;
 12     __int64 nSum;
 13     __int64 Mark;
 14 }segTree[MAXN*4];
 15 __int64 num[MAXN];   //__int64是Windows专用的专门用于记录超大数据的类型，防止越界
 16 
 17 
 18 
 19 void Build(int index,int l,int r)     //建树
 20 {
 21     segTree[index].left=l;
 22     segTree[index].right=r;
 23     if(segTree[index].left==segTree[index].right)
 24     {
 25         segTree[index].nSum=num[l-1];
 26         return ;
 27     }
 28     int mid=(l+r)/2;
 29     Build(2*index,l,mid);
 30     Build(2*index+1,mid+1,r);
 31     segTree[index].nSum=segTree[index*2].nSum+segTree[index*2+1].nSum;           //每个节点和
 32 }
 33 
 34 
 35 
 36 
 37 
 38 void Add(int i,int left,int right,int b)
 39 {
 40     if(segTree[i].left>=left&&segTree[i].right<=right)
 41     {
 42         segTree[i].nSum+=(segTree[i].right-segTree[i].left+1)*b;//倍数加
 43         segTree[i].Mark+=b;//用+=原因是某段可能会被多次标记
 44         return ;
 45     }    
 46     else
 47     {
 48         if(segTree[i].Mark)                   //最关键部分就是这个标记下传，将增加值val传给他的左右孩子
 49         {
 50             segTree[i*2].nSum+=segTree[i].Mark*(segTree[i*2].right-segTree[i*2].left+1);
 51             segTree[i*2].Mark+=segTree[i].Mark;
 52             segTree[i*2+1].nSum+=segTree[i].Mark*(segTree[i*2+1].right-segTree[i*2+1].left+1);
 53             segTree[i*2+1].Mark+=segTree[i].Mark;
 54             segTree[i].Mark=0;
 55         }
 56         if(right>=segTree[i*2+1].left)
 57             Add(2*i+1,left,right,b);
 58         if(left<=segTree[i*2].right)
 59             Add(2*i,left,right,b);    
 60 
 61         segTree[i].nSum=segTree[i*2].nSum+segTree[i*2+1].nSum;
 62     }
 63 
 64 
 65 }
 66 
 67 
 68 
 69 __int64 Query(int i,int left,int right)
 70 {
 71     if(segTree[i].left>=left&&segTree[i].right<=right)
 72             return segTree[i].nSum;
 73     if(segTree[i].Mark)         //求和同样需要标记下传
 74         {
 75             segTree[i*2].nSum+=segTree[i].Mark*(segTree[i*2].right-segTree[i*2].left+1);
 76             segTree[i*2].Mark+=segTree[i].Mark;
 77             segTree[i*2+1].nSum+=segTree[i].Mark*(segTree[i*2+1].right-segTree[i*2+1].left+1);
 78             segTree[i*2+1].Mark+=segTree[i].Mark;
 79             segTree[i].Mark=0;
 80         }
 81         int mid=(segTree[i].left+segTree[i].right)/2;
 82         __int64 max1=0,max2=0;
 83     if(mid<left)
 84             max1= Query(2*i+1,left,right);
 85     else
 86         if(right<=mid)
 87                 max1=Query(2*i,left,right);
 88         else
 89             {
 90                 max1=Query(2*i,left,mid);
 91                 max2=Query(2*i+1,mid+1,right);
 92             }
 93     return max1+max2;
 94 
 95 
 96 
 97 }
 98 
 99 
100 
101 
102 
103 int main()
104 {
105     int n,t;
106     int x,y,z;
107 
108     char C;
109     while(scanf("%d%d",&n,&t)!=EOF)
110     {
111         memset(segTree,0,sizeof(segTree));
112         for(int i=0; i<n; i++)
113             scanf("%I64d",&num[i]);
114         Build(1,1,n);
115         for (int j=0;j<t;j++)
116         {
117             getchar();
118             scanf("%c%d%d",&C,&x,&y);
119 
120             if(C == ‘C‘)
121             {
122                 scanf("%d",&z);
123                 Add(1,x,y,z);
124             }
125             else 
126                 printf("%I64d\n",Query(1,x,y));    
127         }
128     
129     }    
130     return 0;
131 }