Priest John’s Busiest Day

题目
司仪必须在婚礼开始或结束时出现，考虑把第$i$场婚礼拆成两个点
$i$：表示司仪在婚礼$i$开始时出现
$i+n$：表示司仪在$i$婚礼结束时出现

然后直接$O(n^2)$建边👇
①
如果$i,j$的婚礼开始时间段有交叉
那么$i$选开始，$j$就只能选结束
②
如果$i$开始时间段和$j$结束时间段有交叉
那么$i$选开始，$j$就只能选开始
③
如果$i$结束时间段和$j$结束时间段有交叉
那么$i$选结束，$j$就只能选开始
④
如果$i$结束和$j$开始有交叉
那么$i$选结束，$j$就只能选结束

因为蒟蒻的双重循环写法问题，$i,j$和$j,i$都会遍历到，所以$i,j$就只考虑单向连边，反边会在$j,i$时判断

code

#include <stack>
#include <queue>
#include <cstdio>
#include <vector>
#include <cstring>
#include <iostream>
using namespace std;
#define MAXN 2005
struct node {int Begin, End;node() {}node( int B, int E ) { Begin = B, End = E; }
}p[MAXN];
queue < int > q;
stack < int > st;
vector < int > G[MAXN], Edge[MAXN];
int n, cnt, tot;
int c[MAXN], du[MAXN], low[MAXN], dfn[MAXN], scc[MAXN], mbe[MAXN];
bool vis[MAXN];void Tarjan( int u ) {dfn[u] = low[u] = ++ cnt;st.push( u );vis[u] = 1;for( int i = 0;i < G[u].size();i ++ ) {int v = G[u][i];if( ! dfn[v] )Tarjan( v ), low[u] = min( low[u], low[v] );else if( vis[v] )low[u] = min( low[u], dfn[v] );}if( low[u] == dfn[u] ) {int v; tot ++;do {v = st.top(); st.pop();scc[v] = tot;vis[v] = 0;} while( v != u );}
}bool check( int i, int j ) {if( p[i].Begin >= p[j].End || p[i].End <= p[j].Begin )return 0;elsereturn 1;
}int main() {scanf( "%d", &n );for( int i = 1;i <= n;i ++ ) {int a, b, c, d, len;scanf( "%d:%d %d:%d %d", &a, &b, &c, &d, &len );p[i] = node( a * 60 + b, a * 60 + b + len );p[i + n] = node( c * 60 + d - len, c * 60 + d );}for( int i = 1;i <= n;i ++ )for( int j = 1;j <= n;j ++ ) {if( i == j ) continue;if( check( i, j ) ) G[i].push_back( j + n );if( check( i, j + n ) ) G[i].push_back( j );if( check( i + n, j ) ) G[i + n].push_back( j + n );if( check( i + n, j + n ) ) G[i + n].push_back( j );}for( int i = 1;i <= ( n << 1 );i ++ )if( ! dfn[i] ) Tarjan( i );for( int i = 1;i <= n;i ++ )if( scc[i] == scc[i + n] ) return ! printf( "NO\n" );else mbe[scc[i]] = scc[i + n], mbe[scc[i + n]] = scc[i];printf( "YES\n" );memset( c, -1, sizeof( c ) );for( int i = 1;i <= ( n << 1 );i ++ )for( int j = 0;j < G[i].size();j ++ )if( scc[i] == scc[G[i][j]] ) continue;else Edge[scc[G[i][j]]].push_back( scc[i] ), du[scc[i]] ++;for( int i = 1;i <= tot;i ++ )if( ! du[i] ) q.push( i );while( ! q.empty() ) {int u = q.front(); q.pop();if( c[u] == -1 ) c[u] = 1, c[mbe[u]] = 0;for( int i = 0;i < Edge[u].size();i ++ ) {int v = Edge[u][i];du[v] --;if( ! du[v] ) q.push( v );}}for( int i = 1;i <= n;i ++ )if( c[scc[i]] )printf( "%.2d:%.2d %.2d:%.2d\n", p[i].Begin / 60, p[i].Begin % 60, p[i].End / 60, p[i].End % 60 );elseprintf( "%.2d:%.2d %.2d:%.2d\n", p[i + n].Begin / 60, p[i + n].Begin % 60, p[i + n].End / 60, p[i + n].End % 60 );return 0;
}

Peaceful Commission

题目
$2i,2i-1$要分开，那么直接彼此建边即可
要求字典序最小，其实发现我们是按顺序$dfs$找解的，是能保证正确性的

code

#include <cstdio>
#include <vector>
#include <cstring>
using namespace std;
#define MAXN 16005
vector < int > G[MAXN];
int n, m, Top;
int st[MAXN];
bool flag[MAXN];bool dfs( int u ) {if( flag[u ^ 1] ) return 0;if( flag[u] ) return 1;flag[u] = 1;st[++ Top] = u;for( int i = 0;i < G[u].size();i ++ )if( ! dfs( G[u][i] ) ) return 0;return 1;
}bool two_sat() {for( int i = 0;i < ( n << 1 );i += 2 )if( ! flag[i] && ! flag[i ^ 1] ) {Top = 0;if( ! dfs( i ) ) {while( Top ) flag[st[Top]] = 0, Top --;if( ! dfs( i ^ 1 ) ) return 0;}}return 1;
}int main() {while( ~ scanf( "%d %d", &n, &m ) ) {memset( flag, 0, sizeof( flag ) );for( int i = 0;i < ( n << 1 );i ++ )G[i].clear();for( int i = 1, a, b;i <= m;i ++ ) {scanf( "%d %d", &a, &b );a --, b --;G[a].push_back( b ^ 1 );G[b].push_back( a ^ 1 );}if( two_sat() ) {for( int i = 0;i < ( n << 1 );i += 2 )if( flag[i] ) printf( "%d\n", i + 1 );else printf( "%d\n", ( i ^ 1 ) + 1 );}elseprintf( "NIE\n" );}return 0;
}

Let’s go home

题目
注意区分‘对’，‘队’的意思哦~
$i$：表示$i$选手留下
$i+n$：表示$i$离开
①队长$i$和两名队员$j,k$
Ⅰ：如果队长离开，那么两位队员必须全部留下
建边$i+n,j$和$i+n,k$
Ⅱ：如果有一名队员离开，那么队长必须留下
注意同队的两名队员之间是不会相互影响的，换言之，一名队长和一名队员留下也是符合条件的
因此不能胡乱建边$(j,k),(j+n,k+n)$
建边$j+n,i$和$k+n,i$

②一对队员
注意两名队员同时离开也是符合题意的，所以不能胡乱建边$(i+n,j),(j+n,i)$
Ⅰ：如果$i$留下，$j$必须离开，建边$i,j+n$
Ⅱ：如果$j$留下，$i$必须离开，建边$j,i+n$

code

#include <stack>
#include <cstdio>
#include <vector>
#include <cstring>
using namespace std;
#define MAXN 30000
stack < int > st;
vector < int > G[MAXN];
int n, T, M, cnt, tot;
int dfn[MAXN], low[MAXN], scc[MAXN];void tarjan( int u ) {dfn[u] = low[u] = ++ cnt;st.push( u );for( int i = 0;i < G[u].size();i ++ ) {int v = G[u][i];if( ! dfn[v] )tarjan( v ), low[u] = min( low[u], low[v] );else if( ! scc[v] )low[u] = min( low[u], dfn[v] );}if( dfn[u] == low[u] ) {int v; ++ tot;do {v = st.top(); st.pop();scc[v] = tot;} while( u != v );}
}int main() {while( ~ scanf( "%d %d", &T, &M ) ) {n = T * 3;cnt = tot = 0;memset( dfn, 0, sizeof( dfn ) );memset( low, 0, sizeof( low ) );memset( scc, 0, sizeof( scc ) );for( int i = 1;i <= ( n << 1 );i ++ )G[i].clear();while( ! st.empty() ) st.pop();for( int i = 1, a, b, c;i <= T;i ++ ) {scanf( "%d %d %d", &a, &b, &c );G[a + n].push_back( b );G[a + n].push_back( c );G[b + n].push_back( a );G[c + n].push_back( a );}for( int i = 1, a, b;i <= M;i ++ ) {scanf( "%d %d", &a, &b );G[a].push_back( b + n );G[b].push_back( a + n );}for( int i = 0;i < ( n << 1 );i ++ )if( ! dfn[i] ) tarjan( i );bool flag = 0;for( int i = 0;i < n;i ++ )if( scc[i] == scc[i + n] ) { flag = 1; break; }if( flag ) printf( "no\n" );else printf( "yes\n" );}return 0;
}

Bomb Game

题目
二分＋2-sat
先把圆能移动的两个边界拆分成两个点，$i$，$i+n$
与上一题做法类似

考虑二分半径，要求不同圆之间不会有交集
也就是半径不会涉及，即两个之间距离要超过$r\ast 2$

在每次二分的半径$mid$中判断是否有交叉
与上一道题建边思想类似，不再赘述

其实就是如果$a,b$矛盾，那就建边$(a,b+n),(b,a+n)$
把矛盾的两个部分错开

code

#include <cmath>
#include <cstdio>
#include <vector>
#include <cstring>
#include <iostream>
using namespace std;
#define eps 1e-5
#define MAXN 300
struct node {int x, y;
}p[MAXN];
vector < int > G[MAXN];
int cnt, tot, n, Top;
int dfn[MAXN], low[MAXN], scc[MAXN], st[MAXN];void tarjan( int u ) {dfn[u] = low[u] = ++ cnt;st[++ Top] = u;for( int i = 0;i < G[u].size();i ++ ) {int v = G[u][i];if( ! dfn[v] )tarjan( v ), low[u] = min( low[u], low[v] );else if( ! scc[v] )low[u] = min( low[u], dfn[v] );}if( dfn[u] == low[u] ) {int v; ++ tot;do {v = st[Top --];scc[v] = tot;} while( u != v && Top );}
}bool check() {for( int i = 1;i <= n;i ++ )if( scc[i] == scc[i + n] ) return 0;return 1;
}double dis( int x1, int y1, int x2, int y2 ) {return sqrt( 1.0 * ( x1 - x2 ) * ( x1 - x2 ) + 1.0 * ( y1 - y2 ) * ( y1 - y2 ) );
}int main() {while( ~ scanf( "%d", &n ) ) {for( int i = 1;i <= n;i ++ )scanf( "%d %d %d %d", &p[i].x, &p[i].y, &p[i + n].x, &p[i + n].y );double l = 0, r = 10000, mid;while( r - l > eps ) {cnt = tot = Top = 0;mid = ( l + r ) / 2;for( int i = 1;i <= ( n << 1 );i ++ )G[i].clear();memset( dfn, 0, sizeof( dfn ) );memset( low, 0, sizeof( low ) );memset( scc, 0, sizeof( scc ) );for( int i = 1;i < n;i ++ )for( int j = i + 1;j <= n;j ++ ) {if( dis( p[i].x, p[i].y, p[j].x, p[j].y ) < mid * 2 ) {G[i].push_back( j + n );G[j].push_back( i + n );}if( dis( p[i].x, p[i].y, p[j + n].x, p[j + n].y ) < mid * 2 ) {G[i].push_back( j );G[j + n].push_back( i + n );}if( dis( p[i + n].x, p[i + n].y, p[j].x, p[j].y ) < mid * 2 ) {G[i + n].push_back( j + n );G[j].push_back( i );}if( dis( p[i + n].x, p[i + n].y, p[j + n].x, p[j + n].y ) < mid * 2 ) {G[i + n].push_back( j );G[j + n].push_back( i );}}for( int i = 1;i <= n;i ++ )if( ! dfn[i] ) tarjan( i );if( check() ) l = mid;else r = mid;}printf( "%.2lf\n", mid );}return 0;
}

Eliminate the Conflict

题目
如果没有$bob$的出拳方式，单单只给$i,j$轮的限制，这道题就会更加水
其实就算加上$bob$的限制也很简单…
对于$bob$的出拳，$Alice$要想不输就必须打平或者赢
提前处理出$A$在$i$轮的两种出拳方式，然后根据后面给的限制，建边即可

code

#include <cstdio>
#include <vector>
#include <cstring>
using namespace std;
#define MAXN 20005
vector < int > G[MAXN];
int T, n, m, cnt, tot, Top;
int dfn[MAXN], low[MAXN], scc[MAXN], st[MAXN];
int bob[MAXN][2];void tarjan( int u ) {dfn[u] = low[u] = ++ cnt;st[++ Top] = u;for( int i = 0;i < G[u].size();i ++ ) {int v = G[u][i];if( ! dfn[v] ) tarjan( v ), low[u] = min( low[u], low[v] );else if( ! scc[v] )low[u] = min( low[u], dfn[v] );}if( dfn[u] == low[u] ) {int v; ++ tot;do {v = st[Top --];scc[v] = tot;} while( u != v && Top );}
}int main() {scanf( "%d", &T );for( int Case = 1;Case <= T;Case ++ ) {memset( dfn, 0, sizeof( dfn ) );memset( low, 0, sizeof( low ) );memset( scc, 0, sizeof( scc ) );cnt = tot = Top = 0;for( int i = 1;i <= ( n << 1 );i ++ )G[i].clear();scanf( "%d %d", &n, &m );for( int i = 1;i <= n;i ++ ) {scanf( "%d", &bob[i][0] );bob[i][0] --;bob[i][1] = ( bob[i][0] + 1 ) % 3;}for( int i = 1, a, b, k;i <= m;i ++ ) {scanf( "%d %d %d", &a, &b, &k );if( ! k ) {if( bob[a][0] != bob[b][0] ) {G[a].push_back( b + n );G[b].push_back( a + n );}if( bob[a][0] != bob[b][1] ) {G[a].push_back( b );G[b + n].push_back( a + n );}if( bob[a][1] != bob[b][0] ) {G[a + n].push_back( b + n );G[b].push_back( a );}if( bob[a][1] != bob[b][1] ) {G[a + n].push_back( b );G[b + n].push_back( a );}}else {if( bob[a][0] == bob[b][0] ) {G[a].push_back( b + n );G[b].push_back( a + n );}if( bob[a][0] == bob[b][1] ) {G[a].push_back( b );G[b + n].push_back( a + n );}if( bob[a][1] == bob[b][0] ) {G[a + n].push_back( b + n );G[b].push_back( a );}if( bob[a][1] == bob[b][1] ) {G[a + n].push_back( b );G[b + n].push_back( a );}}}for( int i = 1;i <= ( n << 1 );i ++ )if( ! dfn[i] ) tarjan( i );bool flag = 1;for( int i = 1;i <= n;i ++ )if( scc[i] == scc[i + n] ) { flag = 0; break; }if( flag ) printf( "Case #%d: yes\n", Case );else printf( "Case #%d: no\n", Case );}return 0;
}

Bit Magic

题目
$i$：表示$a[i]=1$
$i+n$：表示$a[i]=0$

拆分成二进制，每一位来一次2-sat，有一位矛盾就$NO$

对于第$k$位的数，$b[i][j]$二进制上第$k$位有两种情况👇
①$b[i][j]\&(1<<k)==1$，$k$位是$1$

Ⅰ：$i\%2==1\&\&j\%2==1$
那么只要$i,j$中有一个对应的$a$值第$k$位上为$1$
也可以同时为1
所以如果$a[i]$第$k$位不是$1$，$a[j]$就必须$1$
建边$(i+n,j),(j+n,i)$

Ⅱ：$i\%2==0\&\&j\%2==0$
$i,j$必须同时为$1$，建边$(i,j),(j,i)$

Ⅲ：剩下的情况，一奇一偶
这个时候要调用异或的法则了，必须一个为$1$，另一个$0$才能异或出来$1$
建边$(i,j+n),(j,i+n),(i+n,j),(j+n,i)$

②$b[i][j]\&(1<<k)==0$，$k$位是$0$

Ⅰ：$i\%2==1\&\&j\%2==1$
为了满足$k$位或出来的结果为$0$，必须$i,j$都为$0$
建边$(i+n,j+n),(j+n,i+n)$

Ⅱ：$i\%2==0\&\&j\%2==0$
$i,j$只要有一个不为$1$
也可以同时不为1，所以当$i$为0的时候，是不确定$j$的取值，没有矛盾地方，无法进行建边
建边$(i,j+n),(j,i+n)$

Ⅲ：剩下的情况，一奇一偶
要么同时为$1$，同时$0$
建边$(i,j),(j,i),(i+n,j+n),(j+n,i+n)$

code

#include <cstdio>
#include <vector>
#include <cstring>
#include <iostream>
using namespace std;
#define MAXN 1005
vector < int > G[MAXN];
int n, Top, cnt, tot;
int dfn[MAXN], low[MAXN], scc[MAXN], st[MAXN];
int b[MAXN][MAXN];void tarjan( int u ) {dfn[u] = low[u] = ++ cnt;st[++ Top] = u;for( int i = 0;i < G[u].size();i ++ ) {int v = G[u][i];if( ! dfn[v] )tarjan( v ), low[u] = min( low[u], low[v] );else if( ! scc[v] )low[u] = min( low[u], dfn[v] );}if( dfn[u] == low[u] ) {int v; tot ++;do {v = st[Top --];scc[v] = tot;} while( u != v && Top );}
}void init() {cnt = tot = Top = 0;memset( scc, 0, sizeof( scc ) );memset( dfn, 0, sizeof( dfn ) );for( int i = 0;i < ( n << 1 );i ++ )G[i].clear();
}int main() {while( ~ scanf( "%d", &n ) ) {for( int i = 0;i < n;i ++ )for( int j = 0;j < n;j ++ )scanf( "%d", &b[i][j] );bool flag = 0;for( int i = 0;i < n;i ++ ) {for( int j = 0;j < n;j ++ )if( i == j && b[i][j] ) { flag = 1; break; }else if( b[i][j] != b[j][i] ) { flag = 1; break; }if( flag ) break;}if( flag ) { printf( "NO\n" ); break; }for( int k = 0;k < 31;k ++ ) {init();for( int i = 0;i < n;i ++ )for( int j = i + 1;j < n;j ++ ) {int temp = ( b[i][j] & ( 1 << k ) );if( temp ) {if( ( i & 1 ) && ( j & 1 ) ) {G[i + n].push_back( j );G[j + n].push_back( i );}else if( i % 2 == 0 && j % 2 == 0 ) {G[i].push_back( j );G[j].push_back( i );}else {G[i].push_back( j + n );G[i + n].push_back( j );G[j].push_back( i + n );G[j + n].push_back( i );}}else {if( ( i & 1 ) && ( j & 1 ) ) {G[i + n].push_back( j + n );G[j + n].push_back( i + n );}else if( i % 2 == 0 && j % 2 == 0 ) {G[i].push_back( j + n );G[j].push_back( i + n );}else {G[i].push_back( j );G[j].push_back( i );G[i + n].push_back( j + n );G[j + n].push_back( i + n );}}}for( int i = 0;i < ( n << 1 );i ++ )if( ! dfn[i] ) tarjan( i );for( int i = 0;i < n;i ++ )if( scc[i] == scc[i + n] ) {flag = 1;break;}if( flag ) break;}if( flag ) printf( "NO\n" );else printf( "YES\n" );}return 0;
} 

感觉这几道模板都挺水的…怎么回事？？
2-sat就是会建边就会A，凡事建完后直接tarjan