弗洛伊德Floyd求最小环

模板：

#include<bits/stdc++.h>

using namespace std;

const int MAXN = ;
const int INF = 0xffffff0;
int temp,Map[MAXN][MAXN],Dist[MAXN][MAXN],pre[MAXN][MAXN],ans[MAXN*];

void Solve(int i,int j,int k)
{
    temp = ;   //回溯，存储最小环
    while(i != j)
    {
        ans[temp++] = j;
        j = pre[i][j];
    }
    ans[temp++] = i;
    ans[temp++] = k;
}
void Floyd(int N)
{
    for(int i = ; i <= N; ++i)
        for(int j = ; j <= N; ++j)
        {
            Dist[i][j] = Map[i][j];
            pre[i][j] = i;
        }
    int MinCircle = INF;    //最小环
    for(int k = ; k <= N; ++k)
    {
        for(int i = ; i <= N; ++i)
        {
            for(int j = ; j <= N; ++j)
            {
                if(i != j && Dist[i][j] != INF && Map[i][k] != INF && Map[k][j] != INF
                   && Dist[i][j] + Map[i][k] + Map[k][j] < MinCircle)
                {
                    MinCircle = min(MinCircle, Dist[i][j] + Map[i][k] + Map[k][j]);
                    Solve(i,j,k);   //回溯存储最小环
                }
            }
        }

        for(int i = ; i <= N; ++i)
        {
            for(int j = ; j <= N; ++j)
            {
                if(Dist[i][k] != INF && Dist[k][j] != INF &&
                   Dist[i][k] + Dist[k][j] < Dist[i][j])
                {
                    Dist[i][j] = Dist[i][k] + Dist[k][j];
                    pre[i][j] = pre[k][j];  //记录点i到点j的路径上，j前边的点
                }
            }
        }
    }

    if(MinCircle == INF)    //不存在环
    {
        printf("No solution.\n");
        return;
    }
    //如果求出最小环为负的，原图必定存在负环
    for(int i = ;i < temp; ++i)    //输出最小环
        if(i != temp-)
            printf("%d ",ans[i]);
        else
            printf("%d\n",ans[i]);
}

例题：

BZOJ1027: [JSOI2007]合金

思路：给定两个点集A和B，求A中最小的一个子集S，使B中所有的点在S的凸包内部。枚举A点集两点i,j(i可以等于j)若B点集中的所有点都在向量i->j的左侧或线段ij上，就连接一条i->j的单向边，然后Floyd求最小环即可。

#include<bits/stdc++.h>

#define eps 1e-8
using namespace std;
struct node
{
    double x,y,z;
}a[],b[];
double multi(node p1,node p2,node p0)
{
    double x1=p1.x-p0.x;
    double y1=p1.y-p0.y;
    double x2=p2.x-p0.x;
    double y2=p2.y-p0.y;
    return x1*y2-x2*y1;
}
double dis(node p1,node p2)
{
    return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));
}
int d[][];
int main()
{
    int n,m;
    scanf("%d%d",&m,&n);
    for(int i=;i<=m;i++) scanf("%lf%lf%lf",&a[i].x,&a[i].y,&a[i].z);
    for(int i=;i<=n;i++) scanf("%lf%lf%lf",&b[i].x,&b[i].y,&b[i].z);
    int i,j,k;double t;
    memset(d,0x3f,sizeof(d));
    for(i=;i<=m;i++)
        for(j=;j<=m;j++)
        {
            for(k=;k<=n;k++)
            {
                t=multi(a[j],b[k],a[i]);
                if(t<-eps) break;
                if( fabs(t)<eps && dis(a[i],b[k])>dis(a[i],a[j]) )break;
            }
            if(k==n+) d[i][j]=;
        }
    for(k=;k<=m;k++)
        for(i=;i<=m;i++)if(i!=k)
            for(j=;j<=m;j++)if(j!=k)
                if( d[i][j]>d[i][k]+d[k][j] )d[i][j]=d[i][k]+d[k][j];
    int  ans=m+;
    for(i=;i<=m;i++) if(ans>d[i][i])ans=d[i][i];
    if(ans==m+)printf("-1\n");else printf("%d\n",ans);

    return ;
}