Just a Hook
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 23450 Accepted Submission(s): 11742
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.
题解：本来屠夫的钩子是铜的价值是1，然后每次使x到y间的钩子变为银2，或金3；线段树；
错了很多次，原来想着把z设为全局变量，最后答案一直不对，经历了各种曲折。。。
代码：
1 #include<stdio.h> 2 const int MAXN=100010; 3 struct Node{ 4 int l,r; 5 int sum,lazy,val;//val不能全局变量。。。 6 }; 7 Node tree[MAXN<<2]; 8 #define NOW tree[root].sum=tree[root<<1].sum+tree[root<<1|1].sum 9 #define lson root<<1,tree[root].l,mid 10 #define rson root<<1|1,mid+1,tree[root].r 11 void build(int root,int l,int r){ 12 tree[root].l=l; 13 tree[root].r=r; 14 tree[root].lazy=0; 15 tree[root].val=0; 16 if(l==r)tree[root].sum=1; 17 else{ 18 int mid=(l+r)>>1; 19 build(lson); 20 build(rson); 21 NOW; 22 } 23 } 24 void update(int root,int l,int r,int v){ 25 if(l==tree[root].l&&r==tree[root].r){ 26 tree[root].lazy=1; 27 tree[root].val=v; 28 tree[root].sum=(r-l+1)*v; 29 } 30 else{ 31 int mid=(tree[root].l+tree[root].r)>>1; 32 if(tree[root].lazy==1){ 33 tree[root].lazy=0; 34 update(lson,tree[root].val); 35 update(rson,tree[root].val); 36 tree[root].val=0; 37 } 38 if(r<=mid)update(root<<1,l,r,v); 39 else if(l>mid)update(root<<1|1,l,r,v); 40 else{//这个不能少了，代表l,r在tree的两个孩子节点内，也就是找到了l,r的区间； 41 update(root<<1,l,mid,v); 42 update(root<<1|1,mid+1,r,v); 43 } 44 NOW; 45 } 46 } 47 int main(){ 48 int T,N,Q,flot=0; 49 scanf("%d",&T); 50 while(T--){ 51 scanf("%d",&N); 52 scanf("%d",&Q); 53 build(1,1,N); 54 while(Q--){ 55 int x,y,z; 56 scanf("%d%d%d",&x,&y,&z); 57 update(1,x,y,z); 58 } 59 printf("Case %d: The total value of the hook is %d. ",++flot,tree[1].sum); 60 } 61 return 0; 62 }
过了一段时间又写了一遍，增强了自己的理解：
代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<algorithm>
#include<set>
using namespace std;
#define mem(x,y) memset(x,y,sizeof(x))
const int MAXN=100010;
struct Node{
int lazy,v;
};
Node tree[MAXN<<2];
#define ll root<<1
#define rr root<<1|1
#define lson ll,l,mid
#define rson rr,mid+1,r
#define LAZY(x) tree[x].lazy
#define V(x) tree[x].v
int ans;
void pushdown(int root,int x){
if(LAZY(root)){
LAZY(ll)=LAZY(root);
LAZY(rr)=LAZY(root);
V(ll)=LAZY(root)*(x-(x>>1));//这里错了半天。。。。。右节点有时候要多一。。。
V(rr)=LAZY(root)*(x>>1);
LAZY(root)=0;
}
}
void pushup(int root){
V(root)=V(ll)+V(rr);
}
void build(int root,int l,int r){
LAZY(root)=0;
int mid=(l+r)>>1;
if(l==r){
V(root)=1;
return;
}
build(lson);
build(rson);
pushup(root);
}
void update(int root,int l,int r,int L,int R,int v){
if(l>=L&&r<=R){
LAZY(root)=v;
V(root)=(r-l+1)*v;
return;
}
pushdown(root,r-l+1);
int mid=(l+r)>>1;
/*if(mid>=R)update(lson,L,R,v);
else if(mid<L)update(rson,L,R,v);
else{
update(lson,L,mid,v);
update(rson,mid+1,R,v);
}*///这样写也可以。。。
if(mid>=L)update(lson,L,R,v);
if(mid<R)update(rson,L,R,v);
pushup(root);
}
void query(int root,int l,int r,int L,int R){
int mid=(l+r)>>1;
if(l>=L&&r<=R){
ans+=V(root);
return;
}
pushdown(root,r-l+1);
if(mid>=L)query(lson,L,R);
if(mid<R)query(rson,L,R);
}
int main(){
int T,N,kase=0;
scanf("%d",&T);
while(T--){
scanf("%d",&N);
build(1,1,N);
int q,a,b,c;
scanf("%d",&q);
while(q--){
scanf("%d%d%d",&a,&b,&c);
update(1,1,N,a,b,c);
}
ans=0;
query(1,1,N,1,N);//也可以不询问直接输出tree[1].v
printf("Case %d: The total value of the hook is %d.
",++kase,ans);
}
return 0;
}
map超时；
代码：
#include<iostream>
#include<cstdio>
#include<map>
#include<algorithm>
using namespace std;
map<int,int>mp;
map<int,int>::iterator it1;
const int MAXN=100010;
struct Node{
int s,e,v;
};
Node dt[MAXN];
int main(){
int T,N,q,kase=0;
scanf("%d",&T);
while(T--){
scanf("%d",&N);
for(int i=1;i<=N;i++)mp[i]=1;
scanf("%d",&q);
for(int i=0;i<q;i++)scanf("%d%d%d",&dt[i].s,&dt[i].e,&dt[i].v);
for(int i=0;i<q;i++){
for(it1=mp.lower_bound(dt[i].s);it1!=mp.end();){
if(it1->first<=dt[i].e)it1->second=dt[i].v,it1++;
else break;
}
}
int ans=0;
for(int i=1;i<=N;i++)ans+=mp[i];
printf("Case %d: The total value of the hook is %d.
",++kase,ans);
}
return 0;
}
过了一段时间又写了遍，本想着一遍a的，但是错了几个小时，实在找不出，问了群里面的大神，原来我是x-(x>>1)没有加括号，这就是代码风格的问题了，我的代码风格存在很大的问题，以至于错误了很难找出来，大神指出了几点问题：
1：关于define 的括号问题；
2：关于字母与运算符的缩进问题；
3：关于宏定义的名称问题；都存在一定问题；以后要注意了，参照谷歌的吧
另一种线段树写法：
extern "C++"{ #include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> using namespace std; typedef long long LL; typedef unsigned u; typedef unsigned long long ull; #define mem(x,y) memset(x,y,sizeof(x)) void SI(double &x){scanf("%lf",&x);} void SI(int &x){scanf("%d",&x);} void SI(LL &x){scanf("%lld",&x);} void SI(u &x){scanf("%u",&x);} void SI(ull &x){scanf("%llu",&x);} void SI(char *s){scanf("%s",s);} void PI(int x){printf("%d",x);} void PI(double x){printf("%lf",x);} void PI(LL x){printf("%lld",x);} void PI(u x){printf("%u",x);} void PI(ull x){printf("%llu",x);} void PI(char *s){printf("%s",s);} #define NL puts(""); #define ll root<<1 #define rr root<<1|1 #define lson ll,l,mid #define rson rr,mid+1,r const int INF=0x3f3f3f3f; const int MAXN=100010; int tree[MAXN<<2]; int lazy[MAXN<<2]; } /* void pushup(int root){ tree[root]=tree[ll]+tree[rr]; } void pushdown(int root,int x){ if(lazy[root]){ lazy[ll]=lazy[root]; lazy[rr]=lazy[root]; // tree[ll]=lazy[root]*(mid-l+1); // tree[rr]=lazy[root]*(r-mid); tree[ll]=lazy[root]*(x-(x>>1)); tree[rr]=lazy[root]*(x>>1); lazy[root]=0; } } void build(int root,int l,int r){ int mid=(l+r)>>1; lazy[root]=0; if(l==r){ tree[root]=1;return ; } build(lson); build(rson); pushup(root); } void update(int root,int l,int r,int L,int R,int C){ if(l>=L&&r<=R){ tree[root]=(r-l+1)*C; lazy[root]=C; return ; } int mid=(l+r)>>1; pushdown(root,r-l+1); if(mid>=L)update(lson,L,R,C); if(mid<R)update(rson,L,R,C); pushup(root); } */ void pushdown(int root){ lazy[ll] = lazy[rr] = lazy[root]; lazy[root] = 0; } void update(int root,int l,int r,int L,int R,int C){ if(l >= L && r <= R){ lazy[root] = C; return; } if(lazy[root]) pushdown(root); int mid = (l + r) >> 1; if(mid >= L) update(lson,L,R,C); if(mid < R) update(rson,L,R,C); } int sum(int root,int l,int r){ int mid = (l + r) >> 1; if(lazy[root]) return (r - l + 1) * lazy[root]; return sum(lson) + sum(rson); } int main(){ //assert(false); int T,kase=0; int N,M; SI(T); while(T--){ SI(N); lazy[1]=1; SI(M); int a,b,c; while(M--){ scanf("%d%d%d",&a,&b,&c); update(1,1,N,a,b,c); } printf("Case %d: The total value of the hook is %d. ",++kase,sum(1,1,N)); } return 0; }
java超时了。。。
注意：左右查找都是mid判断的。。。注意lazy要把左右支lazy
代码：
import java.util.Scanner; public class hdoj1698{ public static void main(String[] argv){ Scanner cin = new Scanner(System.in); SegmentTree atree = new SegmentTree(100010 << 2); int T, N, q; T = cin.nextInt(); int kase = 0; while(T-- > 0){ N = cin.nextInt(); atree.build(1, 1, N); q = cin.nextInt(); while(q-- > 0){ int a, b, c; a = cin.nextInt(); b = cin.nextInt(); c = cin.nextInt(); atree.update(1, 1, N, a, b, c); } // System.out.println(atree.tree[1]); System.out.println("Case " + (++kase) + ": The total value of the hook is " + atree.query(1, 1, N, 1, N) + "."); } } } class SegmentTree{ private int[] tree; private int[] lazy; public SegmentTree(int size) { tree = new int[size]; lazy = new int[size]; } private void pushup(int root){ tree[root] = tree[root << 1] + tree[root<<1 | 1]; } private void pushdown(int root, int x){ if(lazy[root] != 0){ lazy[root << 1] = lazy[root << 1|1] = lazy[root]; tree[root << 1] = (x - (x >> 1))*lazy[root]; tree[root << 1|1] = (x >> 1)*lazy[root]; lazy[root] = 0; } } public void build(int root, int l, int r){ lazy[root] = 0; if(l == r){ tree[root] = 1; return ; } int mid = (l + r) >> 1; build(root << 1, l, mid); build(root << 1|1, mid + 1, r); pushup(root); } public void update(int root, int l, int r, int L, int R, int v){ if(l >= L && r <= R){ tree[root] = (r - l + 1)*v; lazy[root] = v; return; } pushdown(root, r - l + 1); int mid = (r + l) >> 1; if(mid >= L){ update(root << 1, l, mid, L, R, v); } if(mid < R){ update(root << 1|1, mid + 1, r, L, R, v); //日了狗了。。。。。。。。。。 } pushup(root); } public int query(int root, int l, int r, int L, int R){ if(l >= L && r <= R){ return tree[root]; } pushdown(root, r - l + 1); int sum = 0; int mid = (l + r) >> 1; if(mid >= L){ sum += query(root << 1, l, mid, L, R); } if(mid < R){ sum += query(root << 1|1, mid +1, r, L, R); } return sum; } }