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#include #include #include #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 inline T max(T x, T y, T z){ return max(max(x, y), z); }template inline T min(T x, T y, T z){ return min(min(x, y), z); }template 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 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\n"); else{ printf("YES\n"); 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\n", S[i] / 60, S[i] % 60, (S[i] + D[i]) / 60, (S[i] + D[i]) % 60); else printf("%02d:%02d %02d:%02d\n", (T[i] - D[i]) / 60, (T[i] - D[i]) % 60, T[i] / 60, T[i] % 60); } } } return 0;}