POJ-3683-Priest John's Busiest Day（2-SAT染色）

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链接：http://poj.org/problem?id=3683
2-SAT求其中一个解，详见2-SAT解法浅析
//#include <bits/stdc++.h>#pragma comment(linker, "/STACK:1024000000,1024000000")#include <cstring>#include <cstdio>#include <algorithm>#include <iostream>#include <queue>#include <cmath>#include <string>#include <map>using namespace std;#define INF 0x3f3f3f3f#define MAXN 2010#define MAXM 2000010int ind,head[MAXN];int low[MAXN],DFN[MAXN],belong[MAXN],top,stack_[MAXN],in_stack[MAXN],deep,cir;struct node{    int v,next;} edge[MAXM];struct no{    int st,ed;} S[MAXN],T[MAXN];void add_(int u, int v){    edge[ind].v=v;    edge[ind].next=head[u];    head[u]=ind++;}void Tarjan(int u){    DFN[u]=low[u]=++deep;    in_stack[u]=1;    stack_[top++]=u;    for(int i=head[u]; i+1; i=edge[i].next)    {        int v=edge[i].v;        if(!DFN[v])        {            Tarjan(v);            low[u]=min(low[u],low[v]);        }        else if(in_stack[v])            low[u]=min(DFN[v],low[u]);    }    int p;    if(low[u]==DFN[u])    {        cir++;        do        {            p=stack_[--top];            in_stack[p]=0;            belong[p]=cir;        }        while(p!=u);    }}bool inv(no x,no y){    if(x.ed<=y.st || y.ed<=x.st)return 0;    return 1;}vector<int> mp[MAXN];int in_[MAXN],clr[MAXN],fir[MAXN];int main(){    int m,n;    while(scanf("%d",&n)!=EOF)    {        ind=deep=top=cir=0;        memset(clr,-1,sizeof(clr));        memset(in_,0,sizeof(in_));        memset(head,-1,sizeof(head));        memset(DFN,0,sizeof(DFN));        memset(in_stack,0,sizeof(in_stack));        for(int i=0; i<=n; ++i)mp[i].clear();        int a,b,c,d,e,f,g;        int cnt=0;        for(int i=0; i<n; ++i)        {            scanf("%d:%d %d:%d %d", &a,&b,&c,&d,&e);            S[cnt].st=a*60+b;            S[cnt].ed=S[cnt].st+e;            cnt++;            S[cnt].ed=c*60+d;            S[cnt].st=S[cnt].ed-e;            cnt++;            for(int j=0; j<i; ++j)            {                if(inv(S[i<<1],S[j<<1]))add_(i<<1,j<<1^1);                if(inv(S[i<<1],S[j<<1^1]))add_(i<<1,j<<1);                if(inv(S[i<<1^1],S[j<<1]))add_(i<<1^1,j<<1^1);                if(inv(S[i<<1^1],S[j<<1^1]))add_(i<<1^1,j<<1);                if(inv(S[j<<1],S[i<<1]))add_(j<<1,i<<1^1);                if(inv(S[j<<1],S[i<<1^1]))add_(j<<1,i<<1);                if(inv(S[j<<1^1],S[i<<1]))add_(j<<1^1,i<<1^1);                if(inv(S[j<<1^1],S[i<<1^1]))add_(j<<1^1,i<<1);            }        }        for(int i=0; i<(n<<1); ++i)            if(!DFN[i])Tarjan(i);        int flag=0;        for(int i=0; i<n; ++i)        {            if(belong[i<<1]==belong[i<<1^1])            {                flag=1;break;            }            fir[ belong[i<<1] ]=belong[i<<1^1];            fir[ belong[i<<1^1] ]=belong[i<<1];        }        if(flag)        {            printf("NO\n");            continue;        }        for(int i=0; i<(n<<1); ++i)        {            for(int j=head[i]; j+1; j=edge[j].next)            {                int v=edge[j].v;                if(belong[v]!=belong[i])                {                    in_[ belong[i] ]++;                    mp[ belong[v] ].push_back( belong[i] );                }            }        }        queue<int> q;        for(int i=1; i<=cir; ++i)            if(!in_[i])q.push(i);        while(!q.empty())        {            int u=q.front();            q.pop();            if(clr[u]==-1)            {                clr[u]=1;                clr[fir[u]]=0;            }            int siz=mp[u].size();            for(int i=0; i<siz; ++i)            {                in_[ mp[u][i] ]--;                if(!in_[ mp[u][i] ])q.push(mp[u][i]);            }        }        printf("YES\n");        for(int i=0; i<(n<<1); i+=2)        {            if(clr[belong[i]]==1)            {                a=S[i].st/60;                b=S[i].st%60;                c=S[i].ed/60;                d=S[i].ed%60;                printf("%02d:%02d %02d:%02d\n",a,b,c,d);            }            else            {                a=S[i^1].st/60;                b=S[i^1].st%60;                c=S[i^1].ed/60;                d=S[i^1].ed%60;                printf("%02d:%02d %02d:%02d\n",a,b,c,d);            }        }    }    return 0;}/*208:00 09:00 3008:15 09:00 20*/
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