poj 3683(2-SAT)

poj 3683
1、区间重合的判定
2、想要输出数最少两位可以用”%.2d”输出
例如数是8，但想输出08，就可以用，如果是14，则还是输出14

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
#define ms(i,j) memset(i, j, sizeof i);

const int MAXN = 1000 + 5;

struct inv
{
    int l, r, time;
}in[MAXN];

struct TwoSAT
{
    vector<int> G[MAXN*2];
    int S[MAXN*2];
    bool mark[MAXN*2];
    int n, c;
    void init(int n)
    {
        this->n = n;
        for (int i=0;i<=n*2;i++)
        {
            G[i].clear();
            mark[i] = false;
        }
    }
    void ins(int u, int v)
    {
        G[u].push_back(v);    
    }
    int dfs(int x)
    {
        if (mark[x^1]) return false;
        if (mark[x]) return true;
        mark[x] = true;
        S[++c] = x;
        for (int i=0;i<G[x].size();i++)
        {
            if (!dfs(G[x][i])) return false;
        }
        return true;
    }
    int solve()
    {
        for (int i=0;i<2*n;i+=2)
        if (!mark[i]&&!mark[i+1])
        {
            c = 0;
            if (!dfs(i))
            {
                while (c>0) mark[S[c--]] = false;
                if (!dfs(i+1)) return false;
            }     
        }
        return true;
    }
}ts;

int n;
int fight(int x1, int y1, int x2, int y2)
{
    if (x2>=x1&&x2<y1) return true;
    if (y2<=y1&&x1<y2) return true;
    return false;
}
void init()
{
    for (int i=0;i<n;i++)
    {
        int h1, h2, m1, m2;
        scanf("%d:%d %d:%d %d", &h1, &m1, &h2, &m2, &in[i].time);
        in[i].l = h1*60+m1;
        in[i].r = h2*60+m2;
    }
}
void solve()//2*x is left, 2*x+1 is right(~)
{
    ts.init(n);
    for (int i=0;i<n;i++)
    for (int j=0;j<n;j++)
    if (i!=j)
    {
        int &x1 = in[i].l, &y1 = in[i].r, &x2 = in[j].l, &y2 = in[j].r, &t1 = in[i].time, &t2 = in[j].time;
        if (fight(x1, x1+t1, x2, x2+t2))//l l
        {
            ts.ins(2*i, 2*j+1);
            ts.ins(2*j, 2*i+1);
        }
        if (fight(x1, x1+t1, y2-t2, y2))//l r
        {
            ts.ins(2*i, 2*j);
            ts.ins(2*j+1, 2*i+1);
        }
        if (fight(y1-t1, y1, y2-t2, y2))//r r
        {
            ts.ins(2*i+1, 2*j);
            ts.ins(2*j+1, 2*i);
        }
        if (fight(y1-t1, y1, x2, x2+t2))//r l
        {
            ts.ins(2*j, 2*i);
            ts.ins(2*i+1, 2*j+1);
        }
    }
    if (ts.solve())
    {
        printf("YES\n");
        for (int i=0;i<n;i++)
        {
            if (ts.mark[i*2])//l
            {
                int tot = in[i].l + in[i].time;
                int h1 = in[i].l / 60, m1 = in[i].l % 60;
                int h2 = tot / 60, m2 = tot % 60;
                printf("%.2d:%.2d %.2d:%.2d\n", h1, m1, h2, m2);
            }
            if (ts.mark[i*2+1])//r
            {
                int tot = in[i].r - in[i].time;
                int h1 = in[i].r / 60, m1 = in[i].r % 60;
                int h2 = tot / 60, m2 = tot % 60;
                printf("%.2d:%.2d %.2d:%.2d\n", h2, m2, h1, m1);
            }
        }
    } else printf("NO\n");
}
int main()
{
    while (scanf("%d", &n)==1)
    {
        init();
        solve();
    }
    return 0;
}