POJ 3683 Priest John's Busiest Day 2-sat输出任意一组解

题目：

http://poj.org/problem?id=3683

题意：

有一个牧师，要去主持婚礼，现在这一天有n场婚礼，每场婚礼都有一个时间段和持续时间，每场婚礼只能在其时间段的开端和末尾举行，问牧师能不能主持所有的婚礼，若能输出YES并输出每场婚礼举办的具体时间，否则输出NO

思路：

每场婚礼时间段的开端和末尾看成点，然后判断任意两场婚礼的开端末尾是否冲突，冲突则建边，主要是判断区间的冲突问题

总结：

这题不难，但让我纠结了好久，写的其中一份代码不知为何就是不对，但稍微改变输出方式就对了，好烦。。。

对于第i场婚礼，i表示婚礼的开端，i+n表示婚礼的末尾，这份代码有点问题，第二份代码没问题

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;

const int N = 2010;
struct edge
{
int to, next;
}g[N*N], g1[N*N];
struct node
{
int st, en;
}arr[N*10];
int cnt, head
, cnt1, head1
;
int num, idx, top, scc
, st
, dfn
, low
, pos
, color
, indeg
;
bool vis
;
int n;
void add_edge(int v, int u)
{
g[cnt].to = u, g[cnt].next = head[v], head[v] = cnt++;
}
void add_edge1(int v, int u)
{
g1[cnt1].to = u, g1[cnt1].next = head1[v], head1[v] = cnt1++;
}
void init()
{
memset(head, -1, sizeof head);
memset(dfn, -1, sizeof dfn);
memset(vis, 0, sizeof vis);
cnt = num = idx = top = 0;
}
void tarjan(int v)
{
dfn[v] = low[v] = ++idx;
vis[v] = true, st[top++] = v;
int u;
for(int i = head[v]; i != -1; i = g[i].next)
{
u = g[i].to;
if(dfn[u] == -1)
{
tarjan(u);
low[v] = min(low[v], low[u]);
}
else if(vis[u]) low[v] = min(low[v], dfn[u]);
}
if(dfn[v] == low[v])
{
num++;
do
{
u = st[--top], vis[u] = false, scc[u] = num;
}while(u != v);
}
}
bool cmp(node a, node b)
{
if(a.en <= b.st || b.en <= a.st) return false;
return true;
}
void toposort()
{
queue<int> que;
for(int i = 1; i <= num; i++)
if(indeg[i] == 0) que.push(i);
while(! que.empty())
{
int v = que.front(); que.pop();
if(color[v] == 0) color[v] = 1, color[pos[v]] = 2;
for(int i = head1[v]; i != -1; i = g1[i].next)
{
int u = g1[i].to;
if(--indeg[u] == 0) que.push(u);
}
}
}
bool solve()
{
for(int i = 1; i <= 2*n; i++)
if(dfn[i] == -1) tarjan(i);
for(int i = 1; i <= n; i++)
{
if(scc[i] == scc[i+n]) return false;
pos[scc[i]] = scc[i+n], pos[scc[i+n]] = scc[i];
}
cnt1 = 0;
memset(head1, -1, sizeof head1);
memset(indeg, 0, sizeof indeg);
memset(color, 0, sizeof color);
for(int i = 1; i <= 2*n; i++)
for(int j = head[i]; j != -1; j = g[j].next)
if(scc[i] != scc[g[j].to])
add_edge1(scc[g[j].to], scc[i]), indeg[scc[i]]++;
toposort();
return true;
}
int main()
{
while(~ scanf("%d", &n))
{
int a, b, c, d, e;
init();
for(int i = 1; i <= n; i++)
{
scanf("%d:%d%d:%d%d", &a, &b, &c, &d, &e);
arr[i].st = a*60 + b, arr[i].en = a*60 + b + e;
arr[i+n].st = c*60 + d - e, arr[i+n].en = c*60 + d;
}
for(int i = 1; i <= n; i++)
for(int j = i + 1; j <= n; j++)
{
if(cmp(arr[i], arr[j])) add_edge(i, j + n), add_edge(j, i + n);
if(cmp(arr[i], arr[j+n])) add_edge(i, j), add_edge(j + n, i + n);
if(cmp(arr[i+n], arr[j])) add_edge(i + n, j + n), add_edge(j, i);
if(cmp(arr[i+n], arr[j+n])) add_edge(i + n, j), add_edge(j + n, i);
}
if(solve())
{//用不注释的这部分输出就错，用注释掉的部分输出就对，实在找不出哪里不对
printf("YES\n");
for(int i = 1; i <= 2*n; i++)
if(color[scc[i]] == 1)
printf("%02d:%02d %02d:%02d\n", arr[i].st/60, arr[i].st%60, arr[i].en/60, arr[i].en%60);
//            for(int i = 1; i <= n; i++)
//                if(color[scc[i]] == 1)
//                    printf("%02d:%02d %02d:%02d\n", arr[i].st/60, arr[i].st%60, arr[i].en/60, arr[i].en%60);
//                else
//                    printf("%02d:%02d %02d:%02d\n", arr[i+n].st/60, arr[i+n].st%60, arr[i+n].en/60, arr[i+n].en%60);
}
else printf("NO\n");
}
return 0;
}

另外一份代码，每场婚礼的开端末尾的点的编号是在一起的，这个就没问题

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;

const int N = 2010;
struct edge
{
int to, next;
}g[N*N], g1[N*N];
struct node
{
int st, en;
}arr[N*10];
int cnt, head
, cnt1, head1
;
int num, idx, top, scc
, st
, dfn
, low
, pos
, color
, indeg
;
bool vis
;
int n;
void add_edge(int v, int u)
{
g[cnt].to = u, g[cnt].next = head[v], head[v] = cnt++;
}
void add_edge1(int v, int u)
{
g1[cnt1].to = u, g1[cnt1].next = head1[v], head1[v] = cnt1++;
}
void init()
{
memset(head, -1, sizeof head);
memset(dfn, -1, sizeof dfn);
memset(vis, 0, sizeof vis);
cnt = num = idx = top = 0;
}
bool cmp(node a, node b)
{
if(a.en <= b.st || b.en <= a.st) return false;
return true;
}
void tarjan(int v)
{
dfn[v] = low[v] = ++idx;
vis[v] = true, st[top++] = v;
int u;
for(int i = head[v]; i != -1; i = g[i].next)
{
u = g[i].to;
if(dfn[u] == -1)
{
tarjan(u);
low[v] = min(low[v], low[u]);
}
else if(vis[u]) low[v] = min(low[v], dfn[u]);
}
if(dfn[v] == low[v])
{
num++;
do
{
u = st[--top], vis[u] = false, scc[u] = num;
}while(u != v);
}
}
void toposort()
{
queue<int> que;
for(int i = 1; i <= num; i++)
if(indeg[i] == 0) que.push(i);
while(! que.empty())
{
int v = que.front(); que.pop();
if(color[v] == 0) color[v] = 1, color[pos[v]] = 2;
for(int i = head1[v]; i != -1; i = g1[i].next)
{
int u = g1[i].to;
if(--indeg[u] == 0) que.push(u);
}
}
}
bool solve()
{
for(int i = 0; i < 2*n; i++)
if(dfn[i] == -1) tarjan(i);
for(int i = 0; i < n; i++)
{
if(scc[2*i] == scc[2*i+1]) return false;
pos[scc[2*i]] = scc[2*i+1], pos[scc[2*i+1]] = scc[2*i];
}
cnt1 = 0;
memset(head1, -1, sizeof head1);
memset(indeg, 0, sizeof indeg);
memset(color, 0, sizeof color);
for(int i = 0; i < 2*n; i++)
for(int j = head[i]; j != -1; j = g[j].next)
if(scc[i] != scc[g[j].to])
add_edge1(scc[g[j].to], scc[i]), indeg[scc[i]]++;
toposort();
return true;
}
int main()
{
while(~ scanf("%d", &n))
{
int a, b, c, d, e;
init();
for(int i = 0; i < n; i++)
{
scanf("%d:%d%d:%d%d", &a, &b, &c, &d, &e);
arr[i*2].st = a*60 + b, arr[i*2].en = a*60 + b + e;
arr[i*2+1].st = c*60 + d - e, arr[i*2+1].en = c*60 + d;
}
for(int i = 0; i < n; i++)
for(int j = i + 1; j < n; j++)
{
if(cmp(arr[2*i], arr[2*j])) add_edge(2*i, 2*j+1), add_edge(2*j, 2*i+1);
if(cmp(arr[2*i], arr[2*j+1])) add_edge(2*i, 2*j), add_edge(2*j+1, 2*i+1);
if(cmp(arr[2*i+1], arr[2*j])) add_edge(2*i+1, 2*j+1), add_edge(2*j, 2*i);
if(cmp(arr[2*i+1], arr[2*j+1])) add_edge(2*i+1, 2*j), add_edge(2*j+1, 2*i);
}
if(solve())
{
printf("YES\n");
for(int i = 0; i < 2*n; i++)
if(color[scc[i]] == 1)
printf("%02d:%02d %02d:%02d\n", arr[i].st/60, arr[i].st%60, arr[i].en/60, arr[i].en%60);
}
else printf("NO\n");
}
return 0;
}