Just a Hook
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 30080 Accepted Submission(s): 14859
Problem Description
In
the game of DotA, Pudge’s meat hook is actually the most horrible thing
for most of the heroes. The hook is made up of several consecutive
metallic sticks which are of the same length.
Now Pudge wants to do some operations on the hook.
Let us number the consecutive metallic sticks of the hook from 1 to N. For each operation, Pudge can change the consecutive metallic sticks, numbered from X to Y, into cupreous sticks, silver sticks or golden sticks.
The total value of the hook is calculated as the sum of values of N metallic sticks. More precisely, the value for each kind of stick is calculated as follows:
For each cupreous stick, the value is 1.
For each silver stick, the value is 2.
For each golden stick, the value is 3.
Pudge wants to know the total value of the hook after performing the operations.
You may consider the original hook is made up of cupreous sticks.
Now Pudge wants to do some operations on the hook.
Let us number the consecutive metallic sticks of the hook from 1 to N. For each operation, Pudge can change the consecutive metallic sticks, numbered from X to Y, into cupreous sticks, silver sticks or golden sticks.
The total value of the hook is calculated as the sum of values of N metallic sticks. More precisely, the value for each kind of stick is calculated as follows:
For each cupreous stick, the value is 1.
For each silver stick, the value is 2.
For each golden stick, the value is 3.
Pudge wants to know the total value of the hook after performing the operations.
You may consider the original hook is made up of cupreous sticks.
Input
The
input consists of several test cases. The first line of the input is
the number of the cases. There are no more than 10 cases.
For each case, the first line contains an integer N, 1<=N<=100,000, which is the number of the sticks of Pudge’s meat hook and the second line contains an integer Q, 0<=Q<=100,000, which is the number of the operations.
Next Q lines, each line contains three integers X, Y, 1<=X<=Y<=N, Z, 1<=Z<=3, which defines an operation: change the sticks numbered from X to Y into the metal kind Z, where Z=1 represents the cupreous kind, Z=2 represents the silver kind and Z=3 represents the golden kind.
For each case, the first line contains an integer N, 1<=N<=100,000, which is the number of the sticks of Pudge’s meat hook and the second line contains an integer Q, 0<=Q<=100,000, which is the number of the operations.
Next Q lines, each line contains three integers X, Y, 1<=X<=Y<=N, Z, 1<=Z<=3, which defines an operation: change the sticks numbered from X to Y into the metal kind Z, where Z=1 represents the cupreous kind, Z=2 represents the silver kind and Z=3 represents the golden kind.
Output
For
each case, print a number in a line representing the total value of the
hook after the operations. Use the format in the example.
Sample Input
1
10
2
1 5 2
5 9 3
Sample Output
Case 1: The total value of the hook is 24.
Source
区间更新
1 #include<iostream> 2 #include<cstdio> 3 #include<cmath> 4 #include<cstring> 5 #include<algorithm> 6 #include<queue> 7 #include<map> 8 #include<set> 9 #include<vector> 10 #include<cstdlib> 11 #include<string> 12 #define eps 0.000000001 13 typedef long long ll; 14 typedef unsigned long long LL; 15 using namespace std; 16 const int N=100000+100; 17 struct node{ 18 int l,r; 19 int val; 20 }tree[N*4]; 21 int flag[N*4]; 22 void pushup(int pos){ 23 tree[pos].val=tree[pos<<1].val+tree[pos<<1|1].val; 24 } 25 void build(int l,int r,int pos){ 26 tree[pos].l=l; 27 tree[pos].r=r; 28 tree[pos].val=1; 29 flag[pos]=0; 30 if(tree[pos].l==tree[pos].r)return; 31 int mid=(tree[pos].l+tree[pos].r)>>1; 32 build(l,mid,pos<<1); 33 build(mid+1,r,pos<<1|1); 34 //pushup(pos); 35 } 36 void update(int l,int r,int x,int pos){ 37 if(tree[pos].l==l&&tree[pos].r==r){ 38 tree[pos].val=x; 39 return; 40 } 41 if(tree[pos].val!=0){ 42 tree[pos<<1].val=tree[pos].val; 43 tree[pos<<1|1].val=tree[pos].val; 44 tree[pos].val=0; 45 } 46 int mid=(tree[pos].l+tree[pos].r)>>1; 47 if(mid>=r)update(l,r,x,pos<<1); 48 else if(mid<l)update(l,r,x,pos<<1|1); 49 else{ 50 update(l,mid,x,pos<<1); 51 update(mid+1,r,x,pos<<1|1); 52 } 53 } 54 int query(int l,int r,int pos){ 55 int mid=(tree[pos].l+tree[pos].r)>>1; 56 if(tree[pos].l==l&&tree[pos].r==r){ 57 if(tree[pos].val){ 58 return tree[pos].val*(tree[pos].r-tree[pos].l+1); 59 } 60 else{ 61 return query(l,mid,pos<<1)+query(mid+1,r,pos<<1|1); 62 } 63 } 64 } 65 int main(){ 66 int Case; 67 scanf("%d",&Case); 68 for(int k=1;k<=Case;k++){ 69 int m,n; 70 scanf("%d",&n); 71 build(1,n,1); 72 scanf("%d",&m); 73 int x,y,z; 74 while(m--){ 75 scanf("%d%d%d",&x,&y,&z); 76 update(x,y,z,1); 77 } 78 int ans=query(1,n,1); 79 printf("Case %d: The total value of the hook is %d. ",k,ans); 80 81 } 82 }