POJ 3683 Priest John's Busiest Day (算竞进阶习题)

2-SAT

可以把每一次仪式看成变量，0/1的取值分别为开头举行和结尾举行。

转换为2-SAT接受的命题，就是看某一次仪式中有没有重合的时间段，有的话，就按照不冲突的形式连有向边。

然后跑tarjan就行啦，我们把时间全部转成分钟方便处理。。

#include <iostream>
#include <cstdio>
#include <stack>
#include <cstring>
#define INF 0x3f3f3f3f
#define full(a, b) memset(a, b, sizeof a)
using namespace std;
typedef long long ll;
inline int lowbit(int x){ return x & (-x); }
inline int read(){
    int X = 0, w = 0; char ch = 0;
    while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }
    while(isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
    return w ? -X : X;
}
inline int gcd(int a, int b){ return a % b ? gcd(b, a % b) : b; }
inline int lcm(int a, int b){ return a / gcd(a, b) * b; }
template<typename T>
inline T max(T x, T y, T z){ return max(max(x, y), z); }
template<typename T>
inline T min(T x, T y, T z){ return min(min(x, y), z); }
template<typename A, typename B, typename C>
inline A fpow(A x, B p, C lyd){
    A ans = 1;
    for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;
    return ans;
}
const int N = 5000;
int n, cnt, k, tot, head[N], S[N], T[N], D[N], dfn[N], low[N], scc[N], val[N];
bool ins[N];
struct Edge { int v, next; } edge[N*N];
stack<int> st;

void addEdge(int a, int b){
    edge[cnt].v = b, edge[cnt].next = head[a], head[a] = cnt ++;
}

bool overlap(int a, int b, int c, int d){
    return (a >= c && a < d) || (b > c && b <= d) || (a <= c && b >= d);
}

void build(){
    while(!st.empty()) st.pop();
    full(head, -1), full(dfn, 0), full(low, 0);
    full(scc, 0), full(val, 0);
    cnt = k = tot = 0;
}

void tarjan(int s){
    dfn[s] = low[s] = ++k;
    ins[s] = true;
    st.push(s);
    for(int i = head[s]; i != -1; i = edge[i].next){
        int u = edge[i].v;
        if(!dfn[u]){
            tarjan(u);
            low[s] = min(low[s], low[u]);
        }
        else if(ins[u]) low[s] = min(low[s], dfn[u]);
    }
    if(dfn[s] == low[s]){
        tot ++;
        int cur;
        do{
            cur = st.top(); st.pop();
            ins[cur] = false;
            scc[cur] = tot;
        }while(cur != s);
    }
}

int main(){

    while(~scanf("%d", &n)){
        build();
        int a, b, c, d;
        for(int i = 1; i <= n; i ++){
            scanf("%d:%d %d:%d %d", &a, &b, &c, &d, &D[i]);
            S[i] = a * 60 + b, T[i] = c * 60 + d;
        }
        for(int i = 1; i < n; i ++){
            for(int j = i + 1; j <= n; j ++){
                if(overlap(S[i], S[i] + D[i], S[j], S[j] + D[j]))
                    addEdge(i, j + n), addEdge(j, i + n);
                if(overlap(S[i], S[i] + D[i], T[j] - D[j], T[j]))
                    addEdge(i, j), addEdge(j + n, i + n);
                if(overlap(T[i] - D[i], T[i], S[j], S[j] + D[j]))
                    addEdge(i + n, j + n), addEdge(j, i);
                if(overlap(T[i] - D[i], T[i], T[j] - D[j], T[j]))
                    addEdge(i + n, j), addEdge(j + n, i);
            }
        }
        for(int i = 1; i <= 2 * n; i ++){
            if(!dfn[i]) tarjan(i);
        }
        bool good = true;
        for(int i = 1; i <= n; i ++){
            if(scc[i] == scc[i + n]){
                good = false;
                break;
            }
        }
        if(!good) printf("NO
");
        else{
            printf("YES
");
            for(int i = 1; i <= n; i ++){
                val[i] = (scc[i] > scc[i + n]);
            }
            for(int i = 1; i <= n; i ++){
                if(!val[i]) printf("%02d:%02d %02d:%02d
", S[i] / 60, S[i] % 60, (S[i] + D[i]) / 60, (S[i] + D[i]) % 60);
                else printf("%02d:%02d %02d:%02d
", (T[i] - D[i]) / 60, (T[i] - D[i]) % 60, T[i] / 60, T[i] % 60);
            }
        }
    }
    return 0;
}