POJ 3683 Priest John's Busiest Day

最新推荐文章于 2020-04-21 09:51:10 发布

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本文详细阐述了2-SAT问题的求解算法，并通过实例代码展示了如何运用结构化方法解决此类问题，包括建立图模型、进行拓扑排序、着色等步骤。

2-SAT经典问题

#include<cstdio>
#include<string.h>
#include<math.h>
#include<stack>
#define N 2010
#define M N*N*3
using namespace std;
struct edge{
	int v,next;
}edge[M];
int n;
int s[N],t[N],seq[N];
int cnt,head1[N],head2[N],head3[N];
int scc,Index,dfn[N],low[N],belong[N];
int top,sstack[N];
bool instack[N],vis[N];
int d[N],col[N];
void addedge(int u,int v,int *head)
{
    edge[cnt].v=v;
    edge[cnt].next=head[u];
    head[u]=cnt++;
}
int MAX(int a,int b){
	return a>b?a:b;
}
int MIN(int a,int b){
	return a>b?b:a;
}
void tarjan(int u)
{
    dfn[u]=low[u]=++Index;
    sstack[++top]=u;
    instack[u]=true;
    for(int j=head1[u];j!=-1;j=edge[j].next)
    {
        int v=edge[j].v;
        if(dfn[v]==0)
        {
            tarjan(v);
            low[u]=MIN(low[v],low[u]);
        }
        else if(instack[v])
        {
            low[u]=MIN(low[u],dfn[v]);
        }
    }
    if(dfn[u]==low[u])
    {
        scc++;
        while(1)
        {
            int tmp=sstack[top--];
			addedge(scc,tmp,head3);
            instack[tmp]=false;
            belong[tmp]=scc;
            if(tmp==u) break;
        }
    }
}

void make_graph(){
	int i,j;
	for(i=1;i<=2*n;i++){
		for(j=head1[i];j!=-1;j=edge[j].next){
			int v=edge[j].v;
			if(belong[i]!=belong[v]){
				addedge(belong[v],belong[i],head2);
			}
		}
	}
}
bool ok(int i,int j){
	if(s[i]>=t[j] || t[i]<=s[j])
		return false;
	return true;
}
bool check(){
	int i;
	for(i=1;i<=n;i++){
		if(belong[i*2]==belong[i*2-1])
			return false;
	}
	return true;
}
void topo(){
	int i,j;
	top=0;
	stack<int>sta;
	for(i=1;i<=scc;i++)
		for(j=head2[i];j!=-1;j=edge[j].next)
			d[edge[j].v]++;
	for(i=1;i<=scc;i++){
		if(!d[i])
			sta.push(i);
	}
	while(!sta.empty()){
		int u=sta.top();
		sta.pop();
		seq[++top]=u;
		for(i=head2[u];i!=-1;i=edge[i].next){
			int v=edge[i].v;
			d[v]--;
			if(!d[v])
				sta.push(v);
		}
	}
}
void paint(int u){
    int k;
    for(k=head2[u];k!=-1;k=edge[k].next)
        if(col[edge[k].v]==0){
            col[edge[k].v]=-1;
            paint(edge[k].v);
        }
}
void colour(){
	int i,j,k;
	for(j=1;j<=top;j++){
		int u=seq[j];
		if(col[u]==0){
			col[u]=1;
			for(i=head3[u];i!=-1;i=edge[i].next){  //枚举连通分量u中的点
				int v=edge[i].v;
				if(col[belong[((v-1)^1)+1]]==0){
                    col[belong[((v-1)^1)+1]]=-1;
                    paint(belong[((v-1)^1)+1]);   //paint函数表示把新建的图中(belong[((v-1)^1)+1)的前向节点删去
				}
			}
		}
	}
}
void printf_path(){
	int i,j;
	for(i=1;i<=n;i++){
		if(col[belong[i*2]]==1){
			printf("%02d:%02d %02d:%02d\n",s[i*2]/60,s[i*2]%60,t[i*2]/60,t[i*2]%60);
		}
		else{
			printf("%02d:%02d %02d:%02d\n",s[i*2-1]/60,s[i*2-1]%60,t[i*2-1]/60,t[i*2-1]%60);
		}
	}
}
int main(){
	int i,j;
	scanf("%d",&n);
	memset(head1,-1,sizeof(head1));cnt=0;
	memset(head2,-1,sizeof(head2));
	memset(head3,-1,sizeof(head3));
	for(i=1;i<=n;i++){
		int u,v,tem;
		scanf("%d:%d",&u,&v);
		s[2*i-1]=u*60+v;
		scanf("%d:%d",&u,&v);
		t[2*i]=u*60+v;
		scanf("%d",&tem);
		t[2*i-1]=s[2*i-1]+tem;
		s[2*i]=t[2*i]-tem;
	}
	for(i=1;i<=n;i++){
		for(j=i+1;j<=n;j++){
			if(ok(2*i,2*j)){
				addedge(2*i,2*j-1,head1);
				addedge(2*j,2*i-1,head1);
			}
			if(ok(2*i-1,2*j)){
				addedge(2*i-1,2*j-1,head1);
				addedge(2*j,2*i,head1);
			}
			if(ok(2*i,2*j-1)){
				addedge(2*i,2*j,head1);
				addedge(2*j-1,2*i-1,head1);
			}
			if(ok(2*i-1,2*j-1)){
				addedge(2*i-1,2*j,head1);
				addedge(2*j-1,2*i,head1);
			}
		}
	}
	scc=0;
	for(i=1;i<=2*n;i++)
		if(!dfn[i])
			tarjan(i);
	if(!check()){
		printf("NO\n");
	}
	else{
		printf("YES\n");
		make_graph();
		topo();
		colour();
		printf_path();
	}
}