Just a Hook

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.  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.  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.

#7691818 zhuyinghui's solution for [Problem C]

Status Accepted Time 936ms Memory 3628kB Length 1374 Lang G++ Submitted 2016-12-08 16:02:18 Shared Select Code #include <iostream> #include <cstdio> #include <cstring> #define maxn 100005 using namespace std; int lazy[maxn<<2],sum[maxn<<2]; void pushup(int rt) { sum[rt]=sum[rt<<1]+sum[rt<<1|1]; } void pushdown(int rt,int p) { if(lazy[rt]) { lazy[rt<<1]=lazy[rt]; lazy[rt<<1|1]=lazy[rt]; sum[rt<<1]=lazy[rt]*(p-(p>>1)); sum[rt<<1|1]=lazy[rt]*(p>>1); lazy[rt]=0; } } void build(int rt,int l,int r) { lazy[rt]=0; sum[rt]=1; //记得这里面要初始化，否则无法正确输出结果，而且赋值为1就可以了，后期递归 的时候会 向向上赋值，或者放在if语句里面 if(l==r) { return ; } build(rt<<1,l,(l+r)>>1); build(rt<<1|1,(l+r)/2+1,r); pushup(rt); } void update(int rt,int l,int r,int x,int y,int z) { if(x<=l&&r<=y) { lazy[rt]=z; //注意所以的都变为等于而不是+= sum[rt]=z*(r-l+1); return ; } pushdown(rt,r-l+1); int mid=(l+r)/2; if(x<=mid) { update(rt<<1,l,mid,x,y,z); } if(y>mid) { update(rt<<1|1,mid+1,r,x,y,z); } pushup(rt); } int main() { int t,n,m,x,y,z; scanf("%d",&t); for(int cou=1;cou<=t;cou++) { scanf("%d %d",&n,&m); build(1,1,n); while(m--) { scanf("%d %d %d",&x,&y,&z); update(1,1,n,x,y,z); } printf("Case %d: The total value of the hook is %d.\n",cou,sum[1]); } return 0; }