2-sat->poj 3683 Priest John's Busiest Day

Priest John's Busiest Day

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 7323 Accepted: 2498 Special Judge

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 S_ito time T_i. 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 D_i 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 S_i toS_i + D_i, or from T_i - D_i to T_i). 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 S_i, T_i and D_i. S_i and T_i 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.

Sample Input

208:00 09:00 3008:15 09:00 20

Sample Output

YES08:00 08:3008:40 09:00

2-sat， 构图， 求强连通分量， 缩点， 拓扑排序， 输出。

#include <iostream>#include <algorithm>#include <string.h>#include <cstdio>#include <string>#include <queue>#include <stack>using namespace std;#define MAXN 2000 + 10string str1, str2;int len;struct Node{int start, end;}temp[MAXN];struct Edge{int u, v, next;}edge[2000 * MAXN], edgetwo[2000 * MAXN];int head[MAXN], e, cnt, ind, n;int headtwo[MAXN], etwo;int dfn[MAXN], low[MAXN], belong[MAXN], cf[MAXN], indegree[MAXN], col[MAXN];bool ins[MAXN];void add(int u, int v){edge[e].u = u;edge[e].v = v;edge[e].next = head[u];head[u] = e++;}void insert(int u, int v){edgetwo[etwo].u = u;edgetwo[etwo].v = v;edgetwo[etwo].next = headtwo[u];headtwo[u] = etwo++;}void init(){e = etwo = cnt = ind = 0;memset(head, -1, sizeof(head));memset(headtwo, -1, sizeof(headtwo));memset(indegree, 0, sizeof(indegree));memset(col, 0, sizeof(col));}int change(int methord){int res = 0;if (methord == 1){res = ((str1[0] - '0') * 10 + str1[1] - '0') * 60 + (str1[3] - '0') * 10 + str1[4] - '0';}else{res = ((str2[0] - '0') * 10 + str2[1] - '0') * 60 + (str2[3] - '0') * 10 + str2[4] - '0';}return res;}string changetwo(int val){string str = "";str += val / 60 / 10 + '0';str += val / 60 % 10 + '0';str += ":";str += val % 60 / 10 + '0';str += val % 60 % 10 + '0';return str;}stack <int> s;void tarjan(int u){low[u] = dfn[u] = ++ind;ins[u] = true;s.push(u);for (int i = head[u]; i != -1; i = edge[i].next){int v = edge[i].v;if (!dfn[v]){tarjan(v);low[u] = min(low[u], low[v]);}else if (ins[v]){low[u] = min(low[u], dfn[v]);}}if (low[u] == dfn[u]){cnt++;int x;do{x = s.top();s.pop();ins[x] = false;belong[x] = cnt;}while (x != u);}}queue <int> q;void torpu(){for (int i = 1; i <= cnt; i++){if (indegree[i] == 0){q.push(i);}}while (!q.empty()){int cur = q.front();q.pop();if (col[cur] == 0){col[cur] = 1;col[cf[cur]] = -1;}for (int i = headtwo[cur]; i != -1; i = edgetwo[i].next){int v = edgetwo[i].v;if (--indegree[v] == 0){q.push(v);}}}}void solve(){for (int i = 0; i < 2 * n; i++){if (!dfn[i]){tarjan(i);}}bool flag = true;for (int i = 0; i < n; i++){if (belong[2 * i] == belong[i * 2 + 1]){flag = false;}cf[belong[2 * i]] = belong[2 * i + 1];cf[belong[2 * i + 1]] = belong[2 * i];}if (!flag){cout << "NO" << endl;}else{cout << "YES" << endl;for (int i = 0; i < e; i++){if (belong[edge[i].u] != belong[edge[i].v]){insert(belong[edge[i].v], belong[edge[i].u]);indegree[belong[edge[i].u]]++;}}torpu();for (int i = 0; i < 2 * n; i++){if (col[belong[i]] == 1){cout << changetwo(temp[i].start) << ' ' << changetwo(temp[i].end) << endl;}}}}void input(){init();cin >> n;for (int i = 0; i < n; i++){cin >> str1 >> str2 >> len;temp[2 * i].start = change(1);temp[2 * i].end = temp[2 * i].start + len;temp[2 * i + 1].end = change(2);temp[2 * i + 1].start = temp[2 * i + 1].end - len; }for (int i = 0; i < 2 * n; i++){for (int j = i + 1 + !(i & 1); j < 2 * n; j++){if (!(temp[i].end <= temp[j].start || temp[i].start >= temp[j].end)){add(i, j ^ 1);add(j, i ^ 1);}}}solve();}int main(){input();return 0;}