Count Color

http://acm.hust.edu.cn:8080/judge/contest/viewProblem.action?pid=45730

Description

Chosen Problem Solving and Program design as an optional course, you are required to solve all kinds of problems. Here, we get a new problem. 

There is a very long board with length L centimeter, L is a positive integer, so we can evenly divide the board into L segments, and they are labeled by 1, 2, ... L from left to right, each is 1 centimeter long. Now we have to color the board - one segment with only one color. We can do following two operations on the board: 

1. "C A B C" Color the board from segment A to segment B with color C. 
2. "P A B" Output the number of different colors painted between segment A and segment B (including). 

In our daily life, we have very few words to describe a color (red, green, blue, yellow…), so you may assume that the total number of different colors T is very small. To make it simple, we express the names of colors as color 1, color 2, ... color T. At the beginning, the board was painted in color 1. Now the rest of problem is left to your. 

Input

First line of input contains L (1 <= L <= 100000), T (1 <= T <= 30) and O (1 <= O <= 100000). Here O denotes the number of operations. Following O lines, each contains "C A B C" or "P A B" (here A, B, C are integers, and A may be larger than B) as an operation defined previously.

Output

Ouput results of the output operation in order, each line contains a number.

Sample Input

2 2 4C 1 1 2P 1 2C 2 2 2P 1 2

Sample Output

21

第一份代码：

#include<stdio.h>#include<algorithm>#include<string.h>#include<iostream>using namespace std;#define MAXN 100010bool flag[MAXN];struct node {int l,r,color;}t[MAXN*4];void make(int st,int ed ,int num){t[num].l=st;t[num].r=ed;t[num].color=1;if(st==ed)return ;int mid=(st+ed)/2;make(st,mid,num*2);make(mid+1,ed,num*2+1);}void updata(int st,int ed,int num,int color){if(t[num].l==st&&t[num].r==ed){t[num].color=color;return ;}if(t[num].color>0){t[num*2].color=t[num].color;t[num*2+1].color=t[num].color;t[num].color=0;}int mid=(t[num].l+t[num].r)/2;if(st>mid)updata(st,ed,num*2+1,color);else if(ed<=mid)updata(st,ed,num*2,color);else {updata(st,mid,num*2,color);updata(mid+1,ed,num*2+1,color);}}void   query(int st,int ed,int num){if(t[num].color>0){flag[t[num].color]=1;return ;}int mid=(t[num].l+t[num].r)/2;if(st>mid)query(st,ed,num*2+1);else if(ed<=mid)query(st,ed,num*2);else {query(st,mid,num*2);query(mid+1,ed,num*2+1);}}int main(){int T,L,O;char str[3];int ans;scanf("%d%d%d",&L,&T,&O);make(1,L,1);while(O--){scanf("%s",str);if(str[0]=='C'){int x,y,clo;scanf("%d%d%d",&x,&y,&clo);updata(x,y,1,clo);}if(str[0]=='P'){int x;int y;memset(flag,0,sizeof(flag));scanf("%d%d",&x,&y);ans=0;query(x,y,1);for(int i=1;i<=30;i++){if(flag[i]==1)ans++;}printf("%d\n",ans);}}return 0;}

第二份代码：

#include<stdio.h>#include<iostream>#include<string.h>using namespace std;struct  {int l,r,color;}g[100005*4];int a[35];void creat(int l,int r,int k){g[k].l=l;g[k].r=r;g[k].color=1;int mid;if(l<r){mid=(l+r)/2;creat(l,mid,k*2);creat(mid+1,r,k*2+1);}}void update(int l,int r,int add,int k){if(g[k].l>=l&&g[k].r<=r){g[k].color=add;return ;}int mid;if(g[k].l<g[k].r){if(g[k].color>0)
//延时标记
{g[k*2].color=g[k*2+1].color=g[k].color;}g[k].color=-1;mid=(g[k].l+g[k].r)/2;if(r<=mid)update(l,r,add,k*2);else if(l>mid)update(l,r,add,k*2+1);else{update(l,mid,add,k*2);update(mid+1,r,add,k*2+1);}}}void quire(int l,int r,int k){if(g[k].color>0)//不用将整个叶子节点全部求出；{a[g[k].color]=1;return;}int mid;if(g[k].l<g[k].r){    mid=(g[k].l+g[k].r)/2;    if(r<=mid)    quire(l,r,k*2);    else if(l>mid)    quire(l,r,k*2+1);    else{    quire(l,mid,k*2);    quire(mid+1,r,k*2+1);}  }}int main(){int m,n,i,j,k,l,r,add;char c;while(scanf("%d%d%d",&m,&n,&k)!=EOF){creat(1,m,1);while(k--){scanf(" %c",&c);if(c=='C'){                scanf("%d%d%d",&l,&r,&add);    if(l>r){l^=r^=l^=r;}update(l,r,add,1);}else{scanf("%d%d",&l,&r);memset(a,0,sizeof(a));if(l>r){l^=r^=l^=r;}quire(l,r,1);int d=0;for(i=1;i<=n;i++){if(a[i]>0)d++;}printf("%d\n",d);}}}return 0;}

第三份代码：

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;#define maxn 100005struct node{int l,r,color;bool flag;}tree[3*maxn];int n,m,tt; void BuildTree(int p,int l,int r){tree[p].l=l; tree[p].r=r;tree[p].color=1; tree[p].flag=false;if(l==r) return;if(l<r){int mid=(l+r)/2;BuildTree(2*p,l,mid);BuildTree(2*p+1,mid+1,r);}//tree[p].color=tree[2*p].color|tree[2*p+1].color;} void change(int p,int l,int r,int x){if(tree[p].l>=l&&tree[p].r<=r){tree[p].color=x;tree[p].flag=true;return ;}if(tree[p].l<tree[p].r){if(tree[p].flag){tree[2*p].color=tree[p].color;tree[2*p].flag=true;tree[2*p+1].color=tree[p].color;tree[2*p+1].flag=true;tree[p].flag=false;}int mid=(tree[p].l+tree[p].r)/2;if(r<=mid)change(2*p,l,r,x);else if(l>mid)change(2*p+1,l,r,x);else{change(2*p,l,mid,x);change(2*p+1,mid+1,r,x);}tree[p].color=tree[2*p].color|tree[2*p+1].color;}}int que(int p,int l,int r){if(tree[p].flag)returntree[p].color;if(tree[p].l==l&&tree[p].r==r)return tree[p].color;if(tree[p].l<tree[p].r){if(tree[p].flag){tree[2*p].flag=true;tree[2*p].color=tree[p].color;tree[2*p+1].flag=true;tree[2*p+1].color=tree[p].color;tree[p].flag=false;}int mid=(tree[p].l+tree[p].r)/2;if(r<=mid)return que(2*p,l,r);else if(l>mid)return que(2*p+1,l,r);elsereturn que(2*p,l,mid)|que(2*p+1,mid+1,r);}} int make_ans(int x){int ans=0;while(x){if(x%2)ans++;x/=2;}return ans;} int main(){int i,j,x,y,d;char c;scanf("%d%d%d",&n,&m,&tt);BuildTree(1,1,n);getchar();while(tt--){scanf("%c",&c);if(c=='C'){scanf("%d%d%d",&x,&y,&d);if(x>y)swap(x,y);change(1,x,y,1<<(d-1));}else if(c=='P'){scanf("%d%d",&x,&y);if(x>y)swap(x,y);int ans=que(1,x,y);printf("%d\n",make_ans(ans));}getchar();}return 0;}