POJ 3683 2-sat 输出解

题意方法和之前的3648基本一样

注意输出解的方式，分成两部分的点集，应该遍历其中一个点集，根据染色判断选择该点集的点还是另一个点集的点

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#define MAXN 5005
#define MAXM 50005
#define INF 1000000000
using namespace std;
int start[MAXN], end[MAXN], last[MAXN];
int n, index, scc;
vector<int>g[MAXN], dag[MAXN];
int dfn[MAXN], low[MAXN], instack[MAXN];
int fa[MAXN], ha[MAXN], color[MAXN], ind[MAXN];
queue<int>q;
stack<int>st;
void init()
{
    index = scc = 0;
    memset(instack, 0, sizeof(instack));
    for(int i = 0; i < MAXN; i++) g[i].clear(), dag[i].clear();
    memset(dfn, 0, sizeof(dfn));
    memset(color, 0, sizeof(color));
    memset(ind, 0, sizeof(ind));
    memset(ha, 0, sizeof(ha));
    while(!q.empty()) q.pop();
    while(!st.empty()) st.pop();
}
bool conflict(int lx, int ly, int rx, int ry)
{
    if(lx >= rx && lx <= ry) return true;
    if(ly >= rx && ly <= ry) return true;
    if(rx >= lx && rx <= ly) return true;
    if(ry >= lx && ry <= ly) return true;
    return false;
}
void build()
{
    for(int i = 1; i <= n; i++)
        for(int j = i + 1; j <= n; j++)
        {
            if(conflict(start[i], start[i] + last[i] - 1, start[j], start[j] + last[j] - 1)) g[i].push_back(j + n), g[j].push_back(i + n);
            if(conflict(start[i], start[i] + last[i] - 1, end[j] - last[j] , end[j] - 1)) g[i].push_back(j), g[j + n].push_back(i + n);
            if(conflict(end[i] - last[i], end[i] - 1, start[j], start[j] + last[j] - 1)) g[i + n].push_back(j + n), g[j].push_back(i);
            if(conflict(end[i] - last[i], end[i] - 1, end[j] - last[j] , end[j] - 1)) g[i + n].push_back(j), g[j + n].push_back(i);
        }
}
void tarjan(int u)
{
    dfn[u] = low[u] = ++index;
    int size = g[u].size();
    int v;
    instack[u] = 1;
    st.push(u);
    for(int i = 0; i < size; i++)
    {
        v = g[u][i];
        if(!dfn[v])
        {
            tarjan(v);
            low[u] = min(low[u], low[v]);
        }
        else if(instack[v]) low[u] = min(low[u], dfn[v]);
    }
    if(dfn[u] == low[u])
    {
        scc++;
        do
        {
            v = st.top();
            st.pop();
            instack[v] = 0;
            fa[v] = scc;
        }while(v != u);
    }
}
void buildDag()
{
    for(int u = 1; u <= 2 * n; u++)
    {
        int size = g[u].size();
        for(int i = 0; i < size; i++)
        {
            int v = g[u][i];
            if(fa[u] != fa[v]) dag[fa[v]].push_back(fa[u]), ind[fa[u]]++;
        }
    }
}
void topsort()
{
    for(int i = 1; i <= scc; i++)
        if(ind[i] == 0) q.push(i);
    while(!q.empty())
    {
        int u = q.front();
        q.pop();
        if(!color[u]) color[u] = 1, color[ha[u]] = 2;
        int size = dag[u].size();
        for(int i = 0; i < size; i++)
        {
            int v = dag[u][i];
            ind[v]--;
            if(ind[v] == 0) q.push(v);
        }
    }
}
void solve()
{
    for(int i = 1; i <= 2 * n; i++)
        if(!dfn[i]) tarjan(i);
    for(int i = 1; i <= n; i++)
        if(fa[i] == fa[i + n])
        {
            puts("NO");
            return;
        }
        else ha[fa[i]] = fa[i + n], ha[fa[i + n]] = fa[i];
    buildDag();
    topsort();
    printf("YES\n");
    for(int i = 1; i <= n; i++)
        if(color[fa[i]] == 1)
        {
            int h1, h2, m1, m2;
            h1 = start[i] / 60;
            h2 = (start[i] + last[i]) / 60;
            m1 = start[i] % 60;
            m2 = (start[i] + last[i]) % 60;
            printf("%02d:%02d %02d:%02d\n", h1, m1, h2, m2);
        }
        else
        {
            int h1, h2, m1, m2;
            h1 = (end[i] - last[i]) / 60;
            h2 = end[i] / 60;
            m1 = (end[i] - last[i]) % 60;
            m2 = end[i] % 60;
            printf("%02d:%02d %02d:%02d\n", h1, m1, h2, m2);
        }
}
int main()
{
    int hour, miniute;
    while(scanf("%d", &n) != EOF)
    {
        init();
        for(int i = 1; i <= n; i++)
        {
            scanf("%d:%d", &hour, &miniute);
            start[i] = hour * 60 + miniute;
            scanf("%d:%d", &hour, &miniute);
            end[i] = hour * 60 + miniute;
            scanf("%d", &last[i]);
        }
        build();
        solve();
    }
    return 0;
}