Gangster
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3123 Accepted Submission(s): 762
Problem Description
There
are two groups of gangsters fighting with each other. The first group
stands in a line, but the other group has a magic gun that can shoot a
range [a, b], and everyone in that range will take a damage of c points.
When a gangster is taking damage, if he has already taken at least P
point of damage, then the damage will be doubled. You are required to
calculate the damage that each gangster in the first group toke.
To simplify the problem, you are given an array A of length N and a magic number P. Initially, all the elements in this array are 0.
Now, you have to perform a sequence of operation. Each operation is represented as (a, b, c), which means: For each A[i] (a <= i <= b), if A[i] < P, then A[i] will be A[i] + c, else A[i] will be A[i] + c * 2.
Compute all the elements in this array when all the operations finish.
To simplify the problem, you are given an array A of length N and a magic number P. Initially, all the elements in this array are 0.
Now, you have to perform a sequence of operation. Each operation is represented as (a, b, c), which means: For each A[i] (a <= i <= b), if A[i] < P, then A[i] will be A[i] + c, else A[i] will be A[i] + c * 2.
Compute all the elements in this array when all the operations finish.
Input
The input consists several testcases.
The first line contains three integers n, m, P (1 <= n, m, P <= 200000), denoting the size of the array, the number of operations and the magic number.
Next m lines represent the operations. Each operation consists of three integers a; b and c (1 <= a <= b <= n, 1 <= c <= 20).
The first line contains three integers n, m, P (1 <= n, m, P <= 200000), denoting the size of the array, the number of operations and the magic number.
Next m lines represent the operations. Each operation consists of three integers a; b and c (1 <= a <= b <= n, 1 <= c <= 20).
Output
Print A[1] to A[n] in one line. All the numbers are separated by a space.
Sample Input
3 2 1
1 2 1
2 3 1
Sample Output
1 3 1
Source
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时间卡，这段时间杭电也卡，我快晕了
线段树，用c++交
1 #include<iostream> 2 #include<cstdio> 3 using namespace std; 4 #define N 200004 5 int n,m,p; 6 struct node 7 { 8 int left,right; 9 int max,min; 10 int num; 11 }tree[N*4]; 12 void build(int l,int r,int pos) 13 { 14 int mid=(l+r)>>1; 15 tree[pos].left=l; 16 tree[pos].right=r; 17 tree[pos].max=0; 18 tree[pos].min=0; 19 tree[pos].num=0; 20 if(l==r) 21 { 22 return ; 23 } 24 build(l,mid,pos<<1); 25 build(mid+1,r,pos<<1|1); 26 } 27 void update(int l,int r,int pos,int v) 28 { 29 int mid=(tree[pos].left+tree[pos].right)>>1; 30 if(tree[pos].left==l&&tree[pos].right==r) 31 { 32 if(tree[pos].min>=p) 33 { 34 tree[pos].min+=v<<1; 35 tree[pos].max+=v<<1; 36 tree[pos].num+=v<<1; 37 return ; 38 } 39 if(tree[pos].max<p) 40 { 41 tree[pos].min+=v; 42 tree[pos].max+=v; 43 tree[pos].num+=v; 44 return ; 45 } 46 } 47 if(tree[pos].num!=0) 48 { 49 tree[pos<<1].num+=tree[pos].num; 50 tree[pos<<1].min+=tree[pos].num; 51 tree[pos<<1].max+=tree[pos].num; 52 tree[pos<<1|1].num+=tree[pos].num; 53 tree[pos<<1|1].min+=tree[pos].num; 54 tree[pos<<1|1].max+=tree[pos].num; 55 tree[pos].num=0; 56 } 57 if(r<=mid) 58 update(l,r,pos<<1,v); 59 else if(l>mid) 60 update(l,r,pos<<1|1,v); 61 else 62 { 63 update(l,mid,pos<<1,v); 64 update(mid+1,r,pos<<1|1,v); 65 } 66 if(tree[pos*2].max>tree[pos<<1|1].max) 67 tree[pos].max=tree[pos<<1].max; 68 else 69 tree[pos].max=tree[pos<<1|1].max; 70 if(tree[pos*2].min>tree[pos<<1|1].min) 71 tree[pos].min=tree[pos<<1|1].min; 72 else 73 tree[pos].min=tree[pos<<1].min; 74 } 75 void query(int pos) 76 { 77 if(tree[pos].left==tree[pos].right) 78 { 79 if(tree[pos].left!=1) 80 printf(" "); 81 printf("%d",tree[pos].num); 82 return; 83 } 84 if(tree[pos].num!=0) 85 { 86 tree[pos<<1].num+=tree[pos].num; 87 tree[pos<<1|1].num+=tree[pos].num; 88 tree[pos].num=0; 89 } 90 query(pos<<1); 91 query(pos<<1|1); 92 } 93 int main() 94 { 95 while(scanf("%d%d%d",&n,&m,&p)>0) 96 { 97 build(1,n,1); 98 while(m--) 99 { 100 int a,b,c; 101 scanf("%d%d%d",&a,&b,&c); 102 update(a,b,1,c); 103 } 104 query(1); 105 printf(" "); 106 } 107 return 0; 108 }