标签：return following note 委员会 problem air 二分图染色 front fill
| Source : POI 2001 | |||
| Time limit : 10 sec | Memory limit : 32 M | ||
Submitted : 2112, Accepted : 641
The Public Peace Commission should be legislated in Parliament of The Democratic Republic of Byteland according to The Very Important Law. Unfortunately one of the obstacles is the fact that some deputies do not get on with some others.
The Commission has to fulfill the following conditions:
Each party has exactly two deputies in the Parliament. All of them are numbered from 1 to 2n. Deputies with numbers 2i-1 and 2i belong to the i-th party .
Task
Write a program, which:
Input
In the first line of the text file SPO.IN there are two non-negative integers n and m. They denote respectively: the number of parties, 1 <= n <= 8000, and the number of pairs of deputies, who do not like each other, 0 <= m <=2 0000. In each of the following m lines there is written one pair of integers a and b, 1 <= a < b <= 2n, separated by a single space. It means that the deputies a and b do not like each other.
There are multiple test cases. Process to end of file.
Output
The text file SPO.OUT should contain one word NIE (means NO in Polish), if the setting up of the Commission is impossible. In case when setting up of the Commission is possible the file SPO.OUT should contain n integers from the interval from 1 to 2n, written in the ascending order, indicating numbers of deputies who can form the Commission. Each of these numbers should be written in a separate line. If the Commission can be formed in various ways, your program may write any of them.
Sample Input
3 2 1 3 2 4
Sample Output
1 4 5
题意同hdu1814
只是这个更简单，无比较字典序
题目大意
根据宪法，Byteland民主共和国的公众和平委员会应该在国会中通过立法程序来创立。 不幸的是，由于某些党派代表之间的不和睦而使得这件事存在障碍。
此委员会必须满足下列条件：
每个党在议会中有2个代表。代表从1编号到2n。 编号为2i-1和2i的代表属于第I个党派。
任务
写一程序：
输入
在文本文件的第一个行有2非负整数n和m。 他们各自表示：党派的数量n，1 < =n < =8000和不友好的代表对m，0 <=m <=20000。 在下面m行的每行为一对整数a,b，1<=a <b<=2n，中间用单个空格隔开。 它们表示代表a,b互相厌恶。
输出
如果委员会不能创立，文本文件中应该包括单词NIE。若能够成立，文本文件SPO.OUT中应该包括n个从区间1到2n选出的整数，按升序写出，每行一个，这些数字为委员会中代表的编号。如果委员会能以多种方法形成，程序可以只写他们的某一个。
样品输入
3 2 1 3 2 4
样品输出
1 4 5
人生第一次2-SAT
/* 2-SAT模板 无比较字典序 Tarjan缩点+拓扑排序+二分图染色 */ #include<iostream> #include<cstdio> #include<cstring> #include<queue> #define maxn 16010 using namespace std; struct node { int next,to; } e[maxn<<4]; node edge[maxn<<4]; int n,m,t,cnt1,top,cnt,tt; int head[maxn],dfn[maxn],low[maxn],belong[maxn],col[maxn],stack[maxn]; bool in_stack[maxn]; int in[maxn],Head[maxn],conflict[maxn],ans[maxn]; void add(int u,int v) { e[t].to=v; e[t].next=head[u]; head[u]=t++; } void add2(int u,int v) { edge[tt].to=v; edge[tt].next=Head[u]; Head[u]=tt++; } void Tarjan(int u) { int t; stack[top++]=u; dfn[u]=low[u]=++cnt; in_stack[u]=true; for(int i=head[u];i!=-1;i=e[i].next) { int v=e[i].to; if(!dfn[v]) { Tarjan(v); low[u]=min(low[u],low[v]); } else if(in_stack[v]) low[u]=min(dfn[v],low[u]); } if(low[u]==dfn[u])//缩点 { cnt1++; do { t=stack[--top]; belong[t]=cnt1;//belong[i]表示属于哪个缩点 in_stack[t]=false; }while(t!=u); } } bool solve() { for(int i=0;i<2*n;i++) if(!dfn[i]) Tarjan(i); for(int i=0; i<2*n; i++) { if(belong[i]==belong[i^1]) return false; conflict[belong[i]]=belong[i^1]; conflict[belong[i^1]]=belong[i];//处理冲突节点 } memset(in,0,sizeof(in)); memset(col,0,sizeof(col)); for(int i=0;i<2*n;i++) { for(int j=head[i];j!=-1;j=e[j].next) { int u=e[j].to; if(belong[i]!=belong[u]) { add2(belong[u],belong[i]);//新建图 in[belong[i]]++; } } } queue<int>q;//对新图进行拓扑 for(int i=1;i<=cnt1;i++) if(in[i]==0) q.push(i); while(!q.empty()) { int u=q.front(); q.pop(); if(col[u]==0)//染色 { col[u]=1; col[conflict[u]]=-1; } for(int i=Head[u];i!=-1;i=edge[i].next) { int v=edge[i].to; in[v]--;//入度 if(in[v]==0) q.push(v); } } memset(ans,0,sizeof(ans)); for(int i=0;i<2*n;i++) if(col[belong[i]]==1) ans[i]=1; return true; } void start() { memset(low,0,sizeof low); memset(dfn,0,sizeof dfn); memset(stack,0,sizeof stack); memset(in_stack,false,sizeof in_stack); memset(conflict,0,sizeof conflict); memset(head,-1,sizeof head); memset(Head,-1,sizeof Head); memset(e,0,sizeof e); memset(edge,0,sizeof edge); memset(belong,0,sizeof belong); cnt=0;top=0;cnt1=0;t=0;tt=0; } int main() { int a,b; while(scanf("%d%d",&n,&m)==2) { start(); while(m--) { scanf("%d%d",&a,&b); a--,b--; add(a,b^1); add(b,a^1); } if(solve()) { for(int i=0;i<2*n;i++) if(ans[i]) printf("%d\n",i+1); } else printf("NIE\n"); } return 0; }
HIT1917Peaceful Commission(2-SAT)
标签：return following note 委员会 problem air 二分图染色 front fill
原文地址：http://www.cnblogs.com/L-Memory/p/6681837.html
