zoukankan      html  css  js  c++  java
  • Parity game

    POJ

    题意:给一个01序列,现在随意拿出来一个区间,然后说出区间内含有奇数个1还是偶数个1,因为有可能存在假话,让你判断前多少条没有假话,也就是查找第一个假话的位置-1.序列长度(n<=1e9),区间数量(m<=10000.)

    分析:边带权和扩展域都能够解决.个人认为扩展域并查集更好理解.

    没时间分析了,留个坑,以后应该也不会回来补的.

    #include<iostream>
    #include<cstdio>
    #include<algorithm>
    #include<cstring>
    #include<cmath>
    #include<queue>
    #include<map>
    #include<set>
    #define ll long long
    using namespace std;
    const int N=20005;
    int b[N],fa[N],d[N];
    struct ppx{int x,y,z;}a[N];
    inline int get(int x){
    	if(x==fa[x])return x;
    	int root=get(fa[x]);
    	d[x]^=d[fa[x]];
    	return fa[x]=root;
    }
    int main(){
    	int n,m,tot=0;scanf("%d%d",&n,&m);
    	for(int i=1;i<=m;++i){
    		char s[5];
    		scanf("%d%d%s",&a[i].x,&a[i].y,s);
    		if(s[0]=='o')a[i].z=1;
    		else a[i].z=0;
    		b[++tot]=a[i].x;b[++tot]=a[i].y;
    	}
    	sort(b+1,b+tot+1);
    	int sum=unique(b+1,b+tot+1)-b-1;
    	for(int i=1;i<=m;++i){
    		a[i].x=lower_bound(b+1,b+sum+1,a[i].x-1)-b;
    		a[i].y=lower_bound(b+1,b+sum+1,a[i].y)-b;
    	}
    	for(int i=1;i<=sum;++i)fa[i]=i;
    	for(int i=1;i<=m;++i){
    		int p=get(a[i].x),q=get(a[i].y);
    		if(p==q){
    			if(d[a[i].x]^d[a[i].y]!=a[i].z){
    				printf("%d
    ",i-1);
    				return 0;
    			}
    		}
    		else{
    			fa[p]=q;d[p]=d[a[i].x]^d[a[i].y]^a[i].z;
    		}
    	}
    	printf("%d
    ",m);
        return 0;
    }
    
    
    #include<iostream>
    #include<cstdio>
    #include<algorithm>
    #include<cstring>
    #include<cmath>
    #include<queue>
    #include<map>
    #include<set>
    #define ll long long
    using namespace std;
    const int N=40005;
    int b[N],fa[N];
    struct ppx{int x,y,z;}a[N];
    inline int get(int x){
    	if(x==fa[x])return x;
    	return fa[x]=get(fa[x]);
    }
    int main(){
    	int n,m,tot=0;scanf("%d%d",&n,&m);
    	for(int i=1;i<=m;++i){
    		char s[5];
    		scanf("%d%d%s",&a[i].x,&a[i].y,s);
    		if(s[0]=='o')a[i].z=1;
    		else a[i].z=0;
    		b[++tot]=a[i].x;b[++tot]=a[i].y;
    	}
    	sort(b+1,b+tot+1);
    	int sum=unique(b+1,b+tot+1)-b-1;
    	for(int i=1;i<=m;++i){
    		a[i].x=lower_bound(b+1,b+sum+1,a[i].x-1)-b;
    		a[i].y=lower_bound(b+1,b+sum+1,a[i].y)-b;
    	}
    	for(int i=1;i<=2*sum;++i)fa[i]=i;
    	for(int i=1;i<=m;++i){
    		int x1=a[i].x,x2=a[i].x+sum;
    		int y1=a[i].y,y2=a[i].y+sum;
    		if(!a[i].z){
    			if(get(x1)==get(y2)){
    				printf("%d
    ",i-1);
    				return 0;
    			}
    			fa[get(x1)]=get(y1);
    			fa[get(x2)]=get(y2);
    		}
    		else{
    			if(get(x1)==get(y1)){
    				printf("%d
    ",i-1);
    				return 0;
    			}
    			fa[get(x1)]=get(y2);
    			fa[get(x2)]=get(y1);
    		}
    	}
    	printf("%d
    ",m);
        return 0;
    }
    
    
  • 相关阅读:
    Fiddler: Creation of interception certificate failed.
    ip地址检查正则表达式 兼容ipv4,ipv6
    母版页与子页的启动过程
    erlang 读取confg文件异常 could not start kernel pid error in config file
    转义字符 显示形式 转换成 实际形式 \\n to \n
    How to use epoll? A complete example in C
    Lex & Flex 词法分析器实践(未完,持续更新)
    我理解的爱情———柳智宇 (转载)
    Learning by doing 系列文章概述
    锁与RCU数据共享机制
  • 原文地址:https://www.cnblogs.com/PPXppx/p/11613407.html
Copyright © 2011-2022 走看看