Description
John is the only priest in his town. September 1st is the John's busiest day in a year because there is an old legend in the town that the couple who get married on that day will be forever blessed by the God of Love. This year N couples plan to get married on the blessed day. The i-th couple plan to hold their wedding from time Si to time Ti. According to the traditions in the town, there must be a special ceremony on which the couple stand before the priest and accept blessings. The i-th couple need Di minutes to finish this ceremony. Moreover, this ceremony must be either at the beginning or the ending of the wedding (i.e. it must be either from Si to Si + Di, or from Ti - Di to Ti). Could you tell John how to arrange his schedule so that he can present at every special ceremonies of the weddings.
Note that John can not be present at two weddings simultaneously.
Input
The first line contains a integer N ( 1 ≤ N ≤ 1000).
The next N lines contain the Si, Ti and Di. Si and Ti are in the format of hh:mm.
Output
The first line of output contains "YES" or "NO" indicating whether John can be present at every special ceremony. If it is "YES", output another N lines describing the staring time and finishing time of all the ceremonies.
题目大意：给n个婚礼，每个婚礼要举办一次祝福，这个祝福只能在婚礼的开始或结束的时候举办（大大的2SAT标志），问能否举办所有祝福，并输出祝福的时间段（任意解）。
思路：可以参考国家集训队2003年伍昱的论文，判冲连边就成。
PS：这题各种不严谨，没说Ti - Si ≥ Di，没保证不会超过24小时，反正都忽视掉是可以AC的，不忽视能不能AC我就不知道了……
1 #include <cstdio> 2 #include <cstring> 3 4 const int MAXN = 1010*2; 5 const int MAXM = MAXN * MAXN * 4; 6 7 struct Topological{ 8 int St[MAXN], c; 9 int n, ecnt, cnt; 10 int head[MAXN], order[MAXN], indeg[MAXN]; 11 int next[MAXM], to[MAXM]; 12 13 void addEdge(int x, int y){ 14 to[ecnt] = y; next[ecnt] = head[x]; head[x] = ecnt++; 15 ++indeg[y]; 16 //printf("%d->%d\n",x,y); 17 } 18 19 void init(int nn){ 20 n = nn; ecnt = 2; 21 memset(head, 0, sizeof(head)); 22 memset(indeg,0,sizeof(indeg)); 23 } 24 25 void build(){ 26 c = cnt = 0; 27 for(int i = 1; i <= n; ++i) 28 if(indeg[i] == 0) St[++c] = i; 29 while(c > 0){ 30 int u = St[c--]; order[cnt++] = u; 31 for(int p = head[u]; p; p = next[p]){ 32 int &v = to[p]; 33 --indeg[v]; 34 if(indeg[v] == 0) St[++c] = v; 35 } 36 } 37 } 38 } T; 39 40 struct TwoSAT{ 41 int St[MAXN], c; 42 int n, ecnt, dfs_clock, scc_cnt; 43 int head[MAXN], sccno[MAXN], pre[MAXN], lowlink[MAXN]; 44 int next[MAXM], to[MAXM]; 45 int select[MAXN], sccnox[MAXN]; 46 47 void dfs(int u){ 48 lowlink[u] = pre[u] = ++dfs_clock; 49 St[++c] = u; 50 for(int p = head[u]; p; p = next[p]){ 51 int &v = to[p]; 52 if(!pre[v]){ 53 dfs(v); 54 if(lowlink[u] > lowlink[v]) lowlink[u] = lowlink[v]; 55 }else if(!sccno[v]){ 56 if(lowlink[u] > pre[v]) lowlink[u] = pre[v]; 57 } 58 } 59 if(lowlink[u] == pre[u]){ 60 sccnox[++scc_cnt] = u; 61 while(true){ 62 int x = St[c--]; 63 sccno[x] = scc_cnt; 64 if(x == u) break; 65 } 66 } 67 } 68 69 void init(int nn){ 70 n = nn; 71 ecnt = 2; dfs_clock = scc_cnt = 0; 72 memset(head,0,sizeof(head)); 73 memset(pre,0,sizeof(pre)); 74 memset(sccno,0,sizeof(sccno)); 75 } 76 77 void addEdge(int x, int y){//x, y clash 78 to[ecnt] = y^1; next[ecnt] = head[x]; head[x] = ecnt++; 79 to[ecnt] = x^1; next[ecnt] = head[y]; head[y] = ecnt++; 80 //printf("%d<>%d\n",x,y); 81 } 82 83 bool solve(){ 84 for(int i = 0; i < n; ++i) 85 if(!pre[i]) dfs(i); 86 for(int i = 0; i < n; i += 2) 87 if(sccno[i] == sccno[i^1]) return false; 88 return true; 89 } 90 91 void bulid_select(){ 92 T.init(scc_cnt); 93 for(int u = 0; u < n; ++u){ 94 for(int p = head[u]; p; p = next[p]){ 95 int &v = to[p]; 96 if(sccno[u] == sccno[v]) continue; 97 T.addEdge(sccno[u], sccno[v]); 98 } 99 } 100 T.build(); 101 memset(select,255,sizeof(select)); 102 for(int i = T.n - 1; i > 0; --i) { 103 int &x = T.order[i]; 104 if(select[x] == -1){ 105 select[x] = 1; 106 select[sccno[sccnox[x]^1]] = 0; 107 } 108 } 109 } 110 } G; 111 112 const int MAXNN = 1010; 113 114 int a1[MAXNN], b1[MAXNN], a2[MAXNN], b2[MAXNN], a3[MAXNN], b3[MAXNN], c[MAXNN]; 115 116 inline bool clash(int beg1, int end1, int beg2, int end2){ 117 if(end1 <= beg2 || end2 <= beg1) return false; 118 return true; 119 } 120 121 int main(){ 122 int n; 123 while(scanf("%d", &n)!=EOF){ 124 G.init(n*2); 125 for(int i = 0; i < n; ++i) { 126 scanf("%d:%d %d:%d %d", &a1[i], &a2[i], &b1[i], &b2[i], &c[i]); 127 a3[i] = a1[i] * 60 + a2[i]; 128 b3[i] = b1[i] * 60 + b2[i]; 129 } 130 for(int i = 0; i < n; ++i) for(int j = i+1; j < n; ++j) if(i != j){ 131 if(clash(a3[i], a3[i] + c[i], a3[j], a3[j] + c[j])) G.addEdge(i*2, j*2); 132 if(clash(a3[i], a3[i] + c[i], b3[j] - c[j], b3[j])) G.addEdge(i*2, j*2+1); 133 if(clash(b3[i] - c[i], b3[i], a3[j], a3[j] + c[j])) G.addEdge(i*2+1, j*2); 134 if(clash(b3[i] - c[i], b3[i], b3[j] - c[j], b3[j])) G.addEdge(i*2+1, j*2+1); 135 } 136 if(G.solve()) printf("YES\n"); 137 else {printf("NO\n"); continue;} 138 G.bulid_select(); 139 for(int i = 0; i < n; ++i){ 140 //printf("%d %d\n",i*2,G.sccno[i*2]); 141 if(G.select[G.sccno[i*2]]){ 142 b1[i] = a1[i]; b2[i] = a2[i] + c[i]; 143 while(b2[i] >= 60) ++b1[i], b2[i] -= 60; 144 } else { 145 a1[i] = b1[i]; a2[i] = b2[i] - c[i]; 146 while(a2[i] < 0) --a1[i], a2[i] += 60; 147 } 148 printf("%02d:%02d %02d:%02d\n", a1[i], a2[i], b1[i], b2[i]); 149 } 150 } 151 return 0; 152 }