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HomeDatabaseMysql TutorialPOJ 3683 Priest John's Busiest Day(2

POJ 3683 Priest John's Busiest Day(2-SAT输出方案) http://poj.org/problem?id=3683 题意: 有N对新人举行婚,且每次婚需要持续d时间,从s时间到t时间之间举行且只能选择s到sd时间或t-d到t时间这两个完整的时间段举行.现在只有一个神父,问他有没有可能参加所

POJ 3683 Priest John's Busiest Day(2-SAT输出方案)

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

题意:

        有N对新人举行婚礼,且每次婚礼需要持续d时间,从s时间到t时间之间举行且只能选择s到s+d时间或t-d到t时间这两个完整的时间段举行.现在只有一个神父,问他有没有可能参加所有新人的婚礼(待完整段时间且任意两对新人的婚礼时间不重叠)? 输出一个可行的方案.

分析:

        每对新人的婚礼时间只有两种选择,直接就可以转化为2-SAT问题.其中如果对于第i个婚礼与第j个婚礼来说:

        假设i先办的时间区间为[a,b]而j后办的时间区间为[c,d],如何判断[a,b]与[c,d]是否发生了冲突呢?(边界相交不算).

        只有下面两种情况下区间[s1,e1]与区间[s2,e2]才规范相交.

        1. s1

        2. s2

        仔细一看上面两种情况是相同的,只要相交的两个区间的e1 e2 > s1 s2 即可保证这两个区间相交.

        (仔细想想上面情况)

        然后对于冲突的每对新人添加边即可.

AC代码:

#include<cstdio>
#include<cstring>
#include<vector>
using namespace std;
const int maxn=1000+10;
struct Time
{
    int s,e,d;//开始,结束,持续
    Time(){}
    Time(int s,int e,int d):s(s),e(e),d(d){}
}t[maxn];
struct TwoSAT
{
    int n;
    vector<int> G[maxn*2];
    int S[maxn*2],c;
    bool mark[maxn*2];

    bool dfs(int x)
    {
        if(mark[x^1]) return false;
        if(mark[x]) return true;
        mark[x]=true;
        S[c++]=x;
        for(int i=0;i<g if return false true void init n this->n=n;
        for(int i=0;i<n g memset void add_clause x xval y yval bool solve for i="0;i<2*n;i+=2)if(!mark[i]" c="0;" if while>0) mark[S[--c]]=false;
                if(!dfs(i+1)) return false;
            }
        }
        return true;
    }
}TS;
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=0;i<n int sh scanf t ts.init for i="0;i<n;i++)" j="i+1;j<n;j++)" if ts.add_clause printf else return><br>
<br>


</n></n></g></int></vector></cstring></cstdio>
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