CEOI 1999 Parity game (并查集+离散化)
来源:互联网 发布:虚拟专用网络 编辑:程序博客网 时间:2024/05/19 17:56
Parity game
Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 7174 Accepted: 2785
Description
Now and then you play the following game with your friend. Your friend writes down a sequence consisting of zeroes and ones. You choose a continuous subsequence (for example the subsequence from the third to the fifth digit inclusively) and ask him, whether this subsequence contains even or odd number of ones. Your friend answers your question and you can ask him about another subsequence and so on. Your task is to guess the entire sequence of numbers.
You suspect some of your friend's answers may not be correct and you want to convict him of falsehood. Thus you have decided to write a program to help you in this matter. The program will receive a series of your questions together with the answers you have received from your friend. The aim of this program is to find the first answer which is provably wrong, i.e. that there exists a sequence satisfying answers to all the previous questions, but no such sequence satisfies this answer.
You suspect some of your friend's answers may not be correct and you want to convict him of falsehood. Thus you have decided to write a program to help you in this matter. The program will receive a series of your questions together with the answers you have received from your friend. The aim of this program is to find the first answer which is provably wrong, i.e. that there exists a sequence satisfying answers to all the previous questions, but no such sequence satisfies this answer.
Input
The first line of input contains one number, which is the length of the sequence of zeroes and ones. This length is less or equal to 1000000000. In the second line, there is one positive integer which is the number of questions asked and answers to them. The number of questions and answers is less or equal to 5000. The remaining lines specify questions and answers. Each line contains one question and the answer to this question: two integers (the position of the first and last digit in the chosen subsequence) and one word which is either `even' or `odd' (the answer, i.e. the parity of the number of ones in the chosen subsequence, where `even' means an even number of ones and `odd' means an odd number).
Output
There is only one line in output containing one integer X. Number X says that there exists a sequence of zeroes and ones satisfying first X parity conditions, but there exists none satisfying X+1 conditions. If there exists a sequence of zeroes and ones satisfying all the given conditions, then number X should be the number of all the questions asked.
Sample Input
1051 2 even3 4 odd5 6 even1 6 even7 10 odd
Sample Output
3
Source
CEOI 1999
[Submit] [Go Back] [Status] [Discuss]
解析:我们用s[i]表示数列前 i 项得和,则有:
若[a,b]区间内 1 的个数为偶数,则s[a-1]与s[b]的奇偶性相同;
若[a,b]区间内 1 的个数为奇数,则s[a-1]与s[b]的奇偶性不同。
我们用f[i]记录与s[i]奇偶性相同的,f[i+x]记录与s[i]奇偶性不同的,因为长度很大,但我们用到的也只是区间端点,最多也就m*2<=1e4,所以要采用离散化。
代码:
#include<cstdio>#include<cstring>#include<cctype>#include<algorithm>using namespace std;const int maxn=5e3;struct tnode{int x,y; bool flag;}p[maxn+20];int q[maxn*2+20];int len,f[maxn*4+20];int getin(){ int ans=0;char tmp; while(!isdigit(tmp=getchar())); do ans=(ans<<3)+(ans<<1)+tmp-'0'; while(isdigit(tmp=getchar())); return ans;}int find(int x){ if(f[x]==x)return x; return f[x]=find(f[x]);}int get(int x){ int l=1,r=q[0],mid; while(l<=r) { mid=(l+r)/2; if(x>q[mid])l=mid+1; else r=mid-1;} return l;}int main(){ int i,j,k,m,x1,x2,y1,y2; char s[10]; k=getin(),m=getin(); for(i=1;i<=m;i++) { p[i].x=getin()-1,p[i].y=getin(); q[2*i-1]=p[i].x,q[2*i]=p[i].y; scanf("%s",s); p[i].flag=(s[0]=='e');} k=m*2,q[0]=1,sort(q+1,q+1+k); for(i=2;i<=k;i++)if(q[i]!=q[i-1])q[++q[0]]=q[i]; for(len=q[0]*2,i=0;i<len;i++)f[i]=i; for(k=1;k<=m;k++) { i=get(p[k].x)-1,j=get(p[k].y)-1; x1=find(i),x2=find((i+q[0])%len); y1=find(j),y2=find((j+q[0])%len); if(p[k].flag) { if(x1==y1)continue; if(x2==y1){printf("%d\n",k-1);return 0;} f[x1]=y1,f[x2]=y2;} else { if(x1==y2)continue; if(x1==y1){printf("%d\n",k-1);return 0;} f[x1]=y2,f[x2]=y1;}} printf("%d\n",m); return 0;}
0 0
- CEOI 1999 Parity game (并查集+离散化)
- Poj 1733 Parity Game(离散化+并查集)
- POJ Parity game 离散化+并查集
- poj 1733 - Parity game(离散化+并查集)
- poj 1733 Parity game 并查集 离散化
- poj1733 Parity game(并查集+离散化--经典)
- pku1733 Parity game(离散化+并查集)
- poj 1733 Parity game 【种类并查集+离散化】
- POJ 1733 parity game (hash离散+并查集)
- poj 1733 Parity game (带权并查集)(离散化)
- poj Parity game(带权并查集)(hash离散化)
- poj 1733 Parity game(离散化+带权并查集+二分查找)
- POJ1733 Parity game(并查集模型+带权并查集+离散化)
- poj 1733 Parity game 【离散+并查集】
- POJ 1733 Parity game(路径压缩并查集+离散化)
- poj 1733 Parity game(带权并查集+离散化)
- POJ 1733 Parity game(离散化+带权并查集)
- POJ 1733Parity game 离散化+带权并查集
- JavaScript 属性描述符
- IDF -PMC与Intel联手演示池化NVMe解决方案
- Android 常用第三方库(长期更新)
- mysql统计表中条目个数的方法举例
- centos7装linux翻译软件
- CEOI 1999 Parity game (并查集+离散化)
- LeetCode|Same Tree-java
- 成为程序员前需要做的10件事
- SELINUX 导致rsync无法同步文件
- [LeetCode-19] Remove Nth Node From End of List(删除倒数第N个节点)
- Maven学习笔记(四)——MyEclipse导入、创建&配置Maven工程
- jeecms各种标签类(大部分,并没有包含一些其他的如text_cut html_cut之类)
- LDAP概念和原理
- RobotFramework安装和配置说明