题意

有一个小镇上只有一个牧师。这个小镇上有一个传说，在九月一日结婚的人会受到爱神的保佑，但是要牧师举办一个仪式。这个仪式要么在婚礼刚刚开始的时候举行，要么举行完婚礼正好结束。
现在已知有n场婚礼，告诉你每一场的开始和结束时间，以及举行仪式所需要的时间。问牧师能否参加所有的婚礼，如果能则输出一种方案。

题解

2-SAT + 路径输出
首先建图不难，判断两个时间段是否相交。
然后按照tarjan缩点，跑一边逆拓扑排序，按照逆序的结果开始选择，并把它的对称点删掉。求出来的就是结果。
逆拓扑排序直接把缩点后的图边反向建立，然后跑一边正的拓扑排序就好了。
注意输出的时候要按照原来婚礼的顺序进行输出。

代码

#include<cstdio>
#include<queue>
#include<cstring>
#include<vector>
#include<queue>
using namespace std;
typedef double db;
typedef long long ll;
typedef unsigned long long ull;
const int nmax = 1e6+7;
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3f;
const ull p = 67;
const ull MOD = 1610612741;
int tot, head[10005], nhead[10005], tot2;
int n;
struct edge{int to, nxt;
}e[nmax<<1], ng[nmax<<1];
void add_edge(int u, int v) {e[tot].to = v;e[tot].nxt = head[u];head[u] = tot++;
}
void add_edge2(int u, int v) {ng[tot2].to = v;ng[tot2].nxt = nhead[u];nhead[u] = tot2++;
}
struct Node{int st1, st2, last;
}node[10005];
int dfn[10005], low[10005], ss[10005], st, scc_cnt, color[10005], in[10005], dfs_clock;
int belong[10005], oppcolor[10005];
bool instack[10005], ban[10005];
vector<int> scc[10005];
queue<int> q;
void init() {tot2 = scc_cnt = st = dfs_clock = tot = 0 ;while(!q.empty()) q.pop();memset(head, -1, sizeof head);memset(nhead, -1, sizeof nhead);memset(dfn, 0, sizeof dfn);memset(low, 0, sizeof low);memset(color, 0, sizeof color);memset(ban, 0 ,sizeof ban);memset(belong, 0, sizeof belong);memset(in, 0, sizeof in);memset(oppcolor, 0, sizeof oppcolor);
}
void tarjan(int u) {dfn[u] = low[u] = ++ dfs_clock;ss[st++] = u;instack[u] = true;for(int i = head[u]; i != -1; i = e[i].nxt) {int v = e[i].to;if(!dfn[v]) {tarjan(v);low[u] = min(low[u], low[v]);} else if(instack[v]) {low[u] = min(low[u], dfn[v]);}}if(low[u] == dfn[u]) {scc_cnt++;scc[scc_cnt].clear();int tmp;while(1) {tmp = ss[--st];color[tmp] = scc_cnt;scc[scc_cnt].push_back(tmp);instack[tmp] = false;if(tmp == u)break;}}
}
void toposort(int u) {q.push(u);in[u] = -1;for(int i = nhead[u]; i != -1; i = ng[i].nxt) {int v = ng[i].to;in[v] --;if(!in[v])toposort(v);}
}
bool notcross(int as, int ae, int bs, int be) {if(be <= as || bs >= ae)return true;elsereturn false;
}
void dfs(int u) {ban[u] = 2;for(int i = nhead[u]; i != -1; i = ng[i].nxt) {int v = ng[i].to;if(ban[v] == 0)dfs(v);}
}
void show(int hour, int minute) {if(hour < 10) printf("0%d:",hour);else printf("%d:",hour);if(minute < 10) printf("0%d",minute);else printf("%d",minute);
}
int main(){while(scanf("%d", &n) != EOF) {init();int sth, stm, edh, edm , last;for(int i = 1; i <= n; ++i) {scanf("%d", &sth);getchar();scanf("%d %d", &stm, &edh);getchar();scanf("%d %d", &edm, &last);node[i].st1 = sth * 60 + stm;node[i].st2 = edh * 60 + edm - last;node[i].last = last;}for(int i = 1; i <= n; ++i) {for(int j = i + 1; j <= n; ++j) {if(!notcross(node[i].st1 , node[i].st1 + node[i].last, node[j].st1, node[j].st1 + node[j].last)) {add_edge(i, j + n);add_edge(j, i + n);}if(!notcross(node[i].st2 ,node[i].st2 + node[i].last, node[j].st1, node[j].st1 + node[j].last)) {add_edge(i + n, j + n);add_edge(j, i);}if(!notcross(node[i].st1, node[i].st1 + node[i].last, node[j].st2, node[j].st2 + node[j].last)) {add_edge(i, j);add_edge(j + n, i + n);}if(!notcross(node[i].st2, node[i].st2 + node[i].last, node[j].st2, node[j].st2+ node[j].last)) {add_edge(i + n, j);add_edge(j + n, i);}}}for(int i = 1; i <= 2* n; ++i) {if(!dfn[i]) {tarjan(i);}}bool isok = true;for(int i = 1; i <= n; ++i) {if(color[i] == color[i + n]) {isok = false;break;}}if(!isok) {printf("NO\n");continue;} elseprintf("YES\n");for(int u = 1; u <= 2 * n; ++u) {for(int i = head[u]; i != -1; i = e[i].nxt) {int v = e[i].to;if(color[u] != color[v]) {add_edge2(color[v], color[u]);in[color[u]] ++;}}if(u <= n) oppcolor[color[u]] = color[u + n];else oppcolor[color[u]] = color[u - n];}for(int i = 1; i <= scc_cnt; ++i) {if(!in[i])toposort(i);}for(int j = 1; j <= scc_cnt; ++j){int now = q.front();q.pop();if(ban[now] == 0) {ban[now] = 1;ban[oppcolor[now]] = 2;for(int i = 0; i < scc[now].size(); ++i) {belong[scc[now][i]] = 1;}}}for(int i = 1; i <= n; ++i) {if(belong[i] == 1) {show(node[i].st1 / 60 , node[i].st1 % 60);printf(" ");show( (node[i].st1 + node[i].last) / 60, (node[i].st1 + node[i].last) % 60);printf("\n");} else {show(node[i].st2 / 60 , node[i].st2 % 60);printf(" ");show( (node[i].st2 + node[i].last) / 60, (node[i].st2 + node[i].last) % 60);printf("\n");}}}return 0;
}

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【算法练习】POJ - 3683 Priest John's Busiest Day （2-SAT）

题意

题解

代码

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