Just a Hook
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9758 Accepted Submission(s): 4776
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.
****************************************/*/*/*/*************************************************************
时间总是不够，并且时间还趋于零散化，以至于我都不得不思考我是不是该晚些睡了，这样就可以有一大段完整的时间，（至少近期来说只是说一说，如果事情越来越多再晚睡吧）有时候也很无语，上篇博客写到一半机房要关门，只好强行发上去当保存了，这里对看过我那篇博客的人说抱歉，等我有时间在补上去。
这道题的代码风格不是模仿学校课件里的，因为我觉得刘汝佳的更好一些。（以上全是废话）注释都写在代码里了。
View Code
1 #include<stdio.h> 2 const int maxn=100050; 3 int sumv[maxn<<2]; 4 int setv[maxn<<2]; 5 void pushup(int rt){///更新当前值 6 sumv[rt]=sumv[rt<<1]+sumv[rt<<1|1]; 7 } 8 void pushdown(int rt,int x){///当此节点改变时，更新此节点，并且传递至下一节点 9 int lc=rt<<1,rc=rt<<1|1,mid=x>>1; 10 if(setv[rt]>=0){ 11 setv[rt<<1]=setv[rt<<1|1]=setv[rt]; 12 sumv[lc]=(x-mid)*setv[rt]; 13 sumv[rc]=mid*setv[rt]; 14 setv[rt]=-1; 15 } 16 } 17 void creat(int l,int r,int rt){///特有的建树，因为已经题中总要初始化 18 setv[rt]=-1; 19 sumv[rt]=1; 20 if(l == r) return ; 21 int mid=(l+r)>>1; 22 creat(l,mid,rt<<1); 23 creat(mid+1,r,rt<<1|1); 24 pushup(rt);///每一个节点在子节点建立完成更新 25 } 26 void update(int l,int r,int x,int L,int R,int rt){///数据更新 27 if(l<=L&&r>=R){///当次节点完全覆盖了要更新区间一部分时在次子树的更新结束 28 setv[rt]=x;///更新进程结束 29 sumv[rt]=x*(R-L+1); 30 return ; 31 } 32 pushdown(rt,R-L+1);///更新数据下移并消除此节点的 33 int mid=(L+R)>>1; 34 if(l<=mid) update(l,r,x,L,mid,rt<<1);///如左边有更新，则更新左边 35 if(r>mid) update(l,r,x,mid+1,R,rt<<1|1);///右边有更新则更新右边 36 37 pushup(rt);///更新此节点 38 } 39 int main() 40 { 41 int t, n,q; 42 int k=1; 43 int a,b,c; 44 scanf("%d",&t); 45 while(t--){ 46 scanf("%d%d",&n,&q); 47 creat(1,n,1); 48 49 for(int i=0;i<q;i++){ 50 scanf("%d%d%d",&a,&b,&c); 51 update(a,b,c,1,n,1); 52 /* for(int j=1;j<20;j++){ 53 printf("a%d = %d ",j,sumv[j]);}*/ 54 } 55 printf("Case %d: The total value of the hook is %d. ",k++,sumv[1]); 56 57 } 58 }