poj3683 2 -sat输出路径

tarjan缩点，拓扑排序染色输出（貌似挑战上面没有拓扑啊，而且这样写还过了= =）

主要是找s,t,d,三者之间的关系，找出合取范式这题就很容易了

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<cstdio>
#include<cassert>
#include<iomanip>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define C 0.5772156649
#define pi acos(-1.0)
#define ll long long
#define mod 1000000007
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#pragma comment(linker, "/STACK:1024000000,1024000000")

using namespace std;

const double g=10.0,eps=1e-;
const int N=+,maxn=+,inf=0x3f3f3f;

stack<int>s;
int ss[N],tt[N],d[N];
int ins[N],inans[N];
int dfn[N],low[N];
int num,index;
vector<int>v[N];
int tarjan(int u)
{
    ins[u]=;
    dfn[u]=low[u]=++index;
    s.push(u);
    for(int i=;i<v[u].size();i++)
    {
        int x=v[u][i];
        if(!dfn[x])
        {
            tarjan(x);
            low[u]=min(low[u],low[x]);
        }
        else if(ins[x]==)low[u]=min(low[u],dfn[x]);
    }
    if(dfn[u]==low[u])
    {
        ++num;
        while(!s.empty()){
            int k=s.top();
            s.pop();
            ins[k]=;
            inans[k]=num;
            if(k==u)break;
        }
    }
}
int main()
{
  /*  ios::sync_with_stdio(false);
    cin.tie(0);*/
    int n;
    scanf("%d",&n);
    memset(ins,,sizeof ins);
    memset(inans,,sizeof inans);
    memset(dfn,,sizeof dfn);
    memset(low,,sizeof low);
    while(!s.empty())s.pop();
    for(int i=;i<=*n;i++)v[i].clear();
    num=index=;
    for(int i=;i<=n;i++)
    {
        char S[],T[];
        scanf("%s%s%d",&S,&T,&d[i]);
        ss[i]=((int)(S[]-'')*+(int)(S[]-''))*+(int)(S[]-'')*+(int)(S[]-'');
        tt[i]=((int)(T[]-'')*+(int)(T[]-''))*+(int)(T[]-'')*+(int)(T[]-'');
    }
    for(int i=;i<=n;i++)
    {
        for(int j=i+;j<=n;j++)
        {
            if(min(ss[i]+d[i],ss[j]+d[j])>max(ss[i],ss[j]))
            {
                v[i].push_back(j+n);
                v[j].push_back(i+n);
            }
            if(min(ss[i]+d[i],tt[j])>max(ss[i],tt[j]-d[j]))
            {
                v[i].push_back(j);
                v[j+n].push_back(i+n);
            }
            if(min(tt[i],ss[j]+d[j])>max(tt[i]-d[i],ss[j]))
            {
                v[i+n].push_back(j+n);
                v[j].push_back(i);
            }
            if(min(tt[i],tt[j])>max(tt[i]-d[i],tt[j]-d[j]))
            {
                v[i+n].push_back(j);
                v[j+n].push_back(i);
            }
        }
    }
    for(int i=;i<=*n;i++)
        if(!dfn[i])
           tarjan(i);
    bool f=;
    for(int i=;i<=n;i++)
        if(inans[i]==inans[i+n])
        {
            f=;
            break;
        }
    if(f)puts("NO");
    else
    {
        puts("YES");
        for(int i=;i<=n;i++)
        {
            if(inans[i]<inans[i+n])printf("%02d:%02d %02d:%02d\n",ss[i]/,ss[i]%,(ss[i]+d[i])/,(ss[i]+d[i])%);
            else printf("%02d:%02d %02d:%02d\n",(tt[i]-d[i])/,(tt[i]-d[i])%,tt[i]/,tt[i]%);
        }
    }
    return ;
}
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