hdu 4046 Panda [线段树]

时间：2015-03-09 18:59:56 阅读：235 评论：0 收藏：0 [点我收藏+]

标签：

Panda

Time Limit: 10000/4000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2565 Accepted Submission(s): 861

Problem Description

When I wrote down this letter, you may have been on the airplane to U.S.
We have known for 15 years, which has exceeded one-fifth of my whole life. I still remember the first time we went to the movies, the first time we went for a walk together. I still remember the smiling face you wore when you were dressing in front of the mirror. I love your smile and your shining eyes. When you are with me, every second is wonderful.
The more expectation I had, the more disappointment I got. You said you would like to go to U.S.I know what you really meant. I respect your decision. Gravitation is not responsible for people falling in love. I will always be your best friend. I know the way is difficult. Every minute thinking of giving up, thinking of the reason why you have held on for so long, just keep going on. Whenever you’re having a bad day, remember this: I LOVE YOU.
I will keep waiting, until you come back. Look into my eyes and you will see what you mean to me.
There are two most fortunate stories in my life: one is finally the time I love you exhausted. the other is that long time ago on a particular day I met you.
Saerdna.

It comes back to several years ago. I still remember your immature face.
The yellowed picture under the table might evoke the countless memory. The boy will keep the last appointment with the girl, miss the heavy rain in those years, miss the love in those years. Having tried to conquer the world, only to find that in the end, you are the world. I want to tell you I didn’t forget. Starry night, I will hold you tightly.

Saerdna loves Panda so much, and also you know that Panda has two colors, black and white.
Saerdna wants to share his love with Panda, so he writes a love letter by just black and white.
The love letter is too long and Panda has not that much time to see the whole letter.
But it‘s easy to read the letter, because Saerdna hides his love in the letter by using the three continuous key words that are white, black and white.
But Panda doesn‘t know how many Saerdna‘s love there are in the letter.
Can you help Panda?

Input

An integer T means the number of test cases T<=100
For each test case:
First line is two integers n, m
n means the length of the letter, m means the query of the Panda. n<=50000,m<=10000
The next line has n characters ‘b‘ or ‘w‘, ‘b‘ means black, ‘w‘ means white.
The next m lines
Each line has two type
Type 0: answer how many love between L and R. (0<=L<=R<n)
Type 1: change the kth character to ch(0<=k<n and ch is ‘b’ or ‘w’)

Output

For each test case, output the case number first.
The answer of the question.

Sample Input

2 5 2 bwbwb 0 0 4 0 1 3 5 5 wbwbw 0 0 4 0 0 2 0 2 4 1 2 b 0 0 4

Sample Output

Case 1: 1 1 Case 2: 2 1 1 0

Source

The 36th ACM/ICPC Asia Regional Beijing Site —— Online Contest

Recommend

lcy | We have carefully selected several similar problems for you: 1166 4045 4043 4041 4042

13075023

2015-03-09 18:29:04

Accepted

4046

1466MS

5532K

4484 B

G++

czy

  1 #include <cstdio>
  2 #include <cstdlib>
  3 #include <cstring>
  4 #define N 50005
  5 
  6 using namespace std;
  7 
  8 int n,m;
  9 int T;
 10 int ccnt;
 11 char s[N];
 12 
 13 typedef struct 
 14 {
 15     int num;
 16     int sw;
 17     int sbw;
 18     int ew;
 19     int ewb;
 20 }PP;
 21 
 22 PP p[4*N];
 23 int mp[4*N];
 24 int ll[4*N],rr[4*N];
 25 
 26 PP build(int i,int l,int r)
 27 {
 28     ll[i]=l;
 29     rr[i]=r;
 30     if(l==r){
 31         mp[l]=i;
 32         p[i].num=0;
 33         if(s[l]==‘b‘) {p[i].ew=p[i].ewb=p[i].sw=p[i].sbw=0;}
 34         else {p[i].sw=1;p[i].sbw=0;p[i].ew=1;p[i].ewb=0;}
 35       //  printf("  i=%d l=%d r=%d num=%d sw=%d sbw=%d ew=%d ewb=%d\n", i,l,r,p[i].num,p[i].sw,p[i].sbw,p[i].ew,p[i].ewb);
 36         return p[i];
 37     }
 38     PP le,ri;
 39     int mid=(l+r)/2;
 40     le=build(i*2,l,mid);
 41     ri=build(i*2+1,mid+1,r);
 42 
 43     p[i].num=le.num+ri.num;
 44     if(le.ew * ri.sbw ==1){
 45         p[i].num++;
 46     }
 47     if(le.ewb * ri.sw ==1){
 48         p[i].num++;
 49     }
 50     p[i].sw=le.sw; 
 51     p[i].ew=ri.ew;
 52     if(p[i].sw==0 && s[l+1]==‘w‘){
 53         p[i].sbw=1;
 54     }
 55     else{
 56         p[i].sbw=0;
 57     }
 58 
 59     if(p[i].ew==0 && s[r-1]==‘w‘){
 60         p[i].ewb=1;
 61     }
 62     else{
 63         p[i].ewb=0;
 64     }
 65   //  printf("  i=%d l=%d r=%d num=%d sw=%d sbw=%d ew=%d ewb=%d\n", i,l,r,p[i].num,p[i].sw,p[i].sbw,p[i].ew,p[i].ewb);
 66     return p[i];
 67 }
 68 
 69 void ini()
 70 {
 71     scanf("%d%d",&n,&m);
 72     scanf("%s",s+1);
 73     //printf("  n=%d m=%d s=%s\n",n,m,s+1 );
 74     build(1,1,n);
 75 //    for(int i=1;i<=9;i++) printf("i=%d num=%d\n",i,p[i].num );
 76 }
 77 
 78 PP query(int i,int l,int r,int L ,int R)
 79 {
 80     PP te;
 81     if(r<L || l>R){
 82         te.num=0;te.sw=te.sbw=te.ew=te.ewb=0;
 83         return te;
 84     }
 85     if(l==r) return p[i];
 86     if(l>=L && r<=R){
 87         //printf("  q i=%d l=%d r=%d num=%d sw=%d sbw=%d ew=%d ewb=%d\n", i,l,r,p[i].num,p[i].sw,p[i].sbw,p[i].ew,p[i].ewb);
 88         return p[i];
 89     }
 90     int mid=(l+r)/2;
 91     PP le,ri;
 92     if(mid>=L){
 93         le=query(i*2,l,mid,L,R);
 94     }
 95     else{
 96         le.num=0;le.sw=le.sbw=le.ew=le.ewb=-2;
 97     }
 98     if(mid<R){
 99         ri=query(i*2+1,mid+1,r,L,R);
100     }
101     else{
102         ri.num=0;ri.sw=ri.sbw=ri.ew=ri.ewb=-2;
103     }
104     te.num=le.num+ri.num;
105     if(le.ew * ri.sbw ==1){
106         te.num++;
107     }
108     if(le.ewb * ri.sw ==1){
109         te.num++;
110     }
111     te.sw=le.sw; 
112     te.ew=ri.ew;
113     if(l+1<=R && te.sw==0 && s[l+1]==‘w‘){
114         te.sbw=1;
115     }
116     else{
117         te.sbw=0;
118     }
119 
120     if(r-1>=L && te.ew==0 && s[r-1]==‘w‘){
121         te.ewb=1;
122     }
123     else{
124         te.ewb=0;
125     }
126     //printf("     le q i=%d l=%d r=%d num=%d sw=%d sbw=%d ew=%d ewb=%d\n", i,l,r,le.num,le.sw,le.sbw,le.ew,le.ewb);
127     //printf("     ri q i=%d l=%d r=%d num=%d sw=%d sbw=%d ew=%d ewb=%d\n", i,l,r,ri.num,ri.sw,ri.sbw,ri.ew,ri.ewb);
128     //printf("  q i=%d l=%d r=%d num=%d sw=%d sbw=%d ew=%d ewb=%d\n", i,l,r,te.num,te.sw,te.sbw,te.ew,te.ewb);
129     return te;
130 }
131 
132 void updata(int i)
133 {
134     int l,r;
135     l=ll[i];
136     r=rr[i];
137     if(i==0) return;
138     PP le,ri;
139     le=p[i*2];ri=p[i*2+1];
140     p[i].num=le.num+ri.num;
141     if(le.ew * ri.sbw ==1){
142         p[i].num++;
143     }
144     if(le.ewb * ri.sw ==1){
145         p[i].num++;
146     }
147     p[i].sw=le.sw; 
148     p[i].ew=ri.ew;
149     if(p[i].sw==0 && s[l+1]==‘w‘){
150         p[i].sbw=1;
151     }
152     else{
153         p[i].sbw=0;
154     }
155 
156     if(p[i].ew==0 && s[r-1]==‘w‘){
157         p[i].ewb=1;
158     }
159     else{
160         p[i].ewb=0;
161     }
162     updata(i/2);
163 }
164 
165 void solve()
166 {
167     int t,L,R;
168     int k;
169     char ch[3];
170     int i,j;
171     PP ans;
172     for(j=1;j<=m;j++){
173         scanf("%d",&t);
174        // printf(" j=%d t=%d\n",j,t );
175         if(t==0){
176             scanf("%d%d",&L,&R);
177            // printf(" L=%d R=%d\n",L,R );
178             ans=query(1,1,n,L+1,R+1);
179             printf("%d\n", ans.num);
180         }
181         else{
182             scanf("%d%s",&k,ch);
183            // printf(" j=%d t=%d k=%d ch=%s\n", j,t,k,ch);
184             k++;
185             s[k]=ch[0];
186             i=mp[k];
187 
188             p[i].num=0;
189             if(s[k]==‘b‘) {p[i].ew=p[i].ewb=p[i].sw=p[i].sbw=0;}
190             else {p[i].sw=1;p[i].sbw=0;p[i].ew=1;p[i].ewb=0;}
191             updata(i/2);
192            // for(int i=1;i<=9;i++) printf("  i=%d  num=%d sw=%d sbw=%d ew=%d ewb=%d\n", i,p[i].num,p[i].sw,p[i].sbw,p[i].ew,p[i].ewb);
193         }
194     }
195 }
196 
197 int main()
198 {
199     scanf("%d",&T);
200     for(ccnt=1;ccnt<=T;ccnt++){
201         ini();
202         printf("Case %d:\n",ccnt );
203         solve();
204     }
205 }

hdu 4046 Panda [线段树]

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原文地址：http://www.cnblogs.com/njczy2010/p/4324222.html

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