zoj 3537 区间dp+计算几何

题意：给定n个点的坐标，先问这些点是否能组成一个凸包，如果是凸包，问用不相交的线来切这个凸包使得凸包只由三角形组成，根据costi, j = |xi + xj| * |yi + yj| % p算切线的费用，问最少的切割费用。

链接：点我

题解：点我

2015-07-20：专题复习

代码稍微修改了一下，顺便发现题号写错了

 #include<cstdio>
 #include<iostream>
 #include<algorithm>
 #include<cstring>
 #include<cmath>
 #include<queue>
 #include<map>
 using namespace std;
 #define MOD 1000000007
 const int INF=0x3f3f3f3f;
 const double eps=1e-;
 typedef long long ll;
 #define cl(a) memset(a,0,sizeof(a))
 #define ts printf("*****\n");
 const int MAXN=;
 int n,m;
 int sgn(double x)
 {
 if(fabs(x) < eps)return ;
 if(x < )return -;
 else return ;
 }
 struct Point
 {
     int x,y;
     Point(){}
 Point(double _x,double _y)
 {
 x = _x;y = _y;
 }
 Point operator -(const Point &b)const
 {
 return Point(x - b.x,y - b.y);
 }
 //叉积
 double operator ^(const Point &b)const
 {
 return x*b.y - y*b.x;
 }
 //点积
 double operator *(const Point &b)const
 {
 return x*b.x + y*b.y;
 }
 //绕原点旋转角度B（弧度值），后x,y的变化
 };
 Point list[MAXN];
 int Stack[MAXN],top;
 double dist(Point a,Point b)
 {
     return sqrt((a-b)*(a-b));
 }
 //相对于list[0]的极角排序
 bool _cmp(Point p1,Point p2)
 {
     double tmp=(p1-list[])^(p2-list[]);
     if(sgn(tmp)>)return true;
     else if(sgn(tmp)== && sgn(dist(p1,list[]) - dist(p2,list[])) <= )
     return true;
     else return false;
 }
 void Graham(int n)
 {
     Point p0;
     int k=;
     p0=list[];
     //找最下边的一个点
     for(int i=;i < n;i++)
     {
         if( (p0.y>list[i].y) || (p0.y==list[i].y && p0.x>list[i].x) )
         {
             p0=list[i];
             k=i;
         }
     }
     swap(list[k],list[]);
     sort(list+,list+n,_cmp);
     if(n==)
     {
         top=;
         Stack[]=;
         return;
     }
     if(n==)
     {
         top=;
         Stack[]=;
         Stack[]=;
         return ;
     }
     Stack[]=;
     Stack[]=;
     top=;
     for(int i=;i < n;i++)
     {
         while(top> && sgn((list[Stack[top-]]-list[Stack[top-]])^(list[i]-list[Stack[top-]])) <= )
         top--;
         Stack[top++]=i;
     }
 }
 int cost[MAXN][MAXN];
 int dis(Point p1,Point p2)//计算题目定义的cost
 {
     return abs(p1.x+p2.x)*abs(p1.y+p2.y)%m;
 }
 int dp[MAXN][MAXN];
 int main()
 {
     int i,j,k;
     #ifndef ONLINE_JUDGE
     freopen("1.in","r",stdin);
     #endif
     while(~scanf("%d%d",&n,&m))
     {
         for(i=;i<n;i++)
         {
             scanf("%d%d",&list[i].x,&list[i].y);
         }
         Graham(n);
         if(top!=n)
         {
             puts("I can't cut.");
             continue;
         }
         cl(cost);
         for(i=;i<n;i++)
             for(j=i+;j<n;j++)
                 cost[i][j]=cost[j][i]=dis(list[i],list[j]);
         for(i=;i<n;i++)
         {
             for(j=i;j<n;j++)dp[i][j]=INF;
             dp[i][(i+)%n]=;
         }
         for(int len=;len<n;len++)
         {
             for(i=;i+len<=n-;i++)
             {
                 j=i+len;
                 for(k=i+;k<=j-;k++)
                 {
                     dp[i][j]=min(dp[i][j],dp[i][k]+dp[k][j]+cost[i][k]+cost[k][j]);
                 }
             }
         }
         /*for(i=n-3;i>=0;i--)
         {
             for(j=i+2;j<n;j++)
             {
                 for(k=i+1;k<=j-1;k++)
                 {
                     dp[i][j]=min(dp[i][j],dp[i][k]+dp[k][j]+cost[i][k]+cost[k][j]);
                 }
             }
         }*/
         printf("%d\n",dp[][n-]);
     }
 }

 #include<cstdio>
 #include<iostream>
 #include<algorithm>
 #include<cstring>
 #include<cmath>
 #include<queue>
 #include<map>
 using namespace std;
 #define MOD 1000000007
 const int INF=0x3f3f3f3f;
 const double eps=1e-;
 typedef long long ll;
 #define cl(a) memset(a,0,sizeof(a))
 #define ts printf("*****\n");
 const int MAXN=;
 int n,m;
 int sgn(double x)
 {
 if(fabs(x) < eps)return ;
 if(x < )return -;
 else return ;
 }
 struct Point
 {
     int x,y;
     Point(){}
 Point(double _x,double _y)
 {
 x = _x;y = _y;
 }
 Point operator -(const Point &b)const
 {
 return Point(x - b.x,y - b.y);
 }
 //叉积
 double operator ^(const Point &b)const
 {
 return x*b.y - y*b.x;
 }
 //点积
 double operator *(const Point &b)const
 {
 return x*b.x + y*b.y;
 }
 //绕原点旋转角度B（弧度值），后x,y的变化
 };
 Point list[MAXN];
 int Stack[MAXN],top;
 double dist(Point a,Point b)
 {
     return sqrt((a-b)*(a-b));
 }
 //相对于list[0]的极角排序
 bool _cmp(Point p1,Point p2)
 {
     double tmp=(p1-list[])^(p2-list[]);
     if(sgn(tmp)>)return true;
     else if(sgn(tmp)== && sgn(dist(p1,list[]) - dist(p2,list[])) <= )
     return true;
     else return false;
 }
 void Graham(int n)
 {
     Point p0;
     int k=;
     p0=list[];
     //找最下边的一个点
     for(int i=;i < n;i++)
     {
         if( (p0.y>list[i].y) || (p0.y==list[i].y && p0.x>list[i].x) )
         {
             p0=list[i];
             k=i;
         }
     }
     swap(list[k],list[]);
     sort(list+,list+n,_cmp);
     if(n==)
     {
         top=;
         Stack[]=;
         return;
     }
     if(n==)
     {
         top=;
         Stack[]=;
         Stack[]=;
         return ;
     }
     Stack[]=;
     Stack[]=;
     top=;
     for(int i=;i < n;i++)
     {
         while(top> && sgn((list[Stack[top-]]-list[Stack[top-]])^(list[i]-list[Stack[top-]])) <= )
         top--;
         Stack[top++]=i;
     }
 }
 int cost[MAXN][MAXN];
 int dis(Point p1,Point p2)//计算题目定义的cost
 {
     return abs(p1.x+p2.x)*abs(p1.y+p2.y)%m;
 }
 int dp[MAXN][MAXN];
 int main()
 {
     int i,j,k;
     #ifndef ONLINE_JUDGE
     freopen("1.in","r",stdin);
     #endif
     while(~scanf("%d%d",&n,&m))
     {
         for(i=;i<n;i++)
         {
             scanf("%d%d",&list[i].x,&list[i].y);
         }
         Graham(n);
         if(top!=n)
         {
             puts("I can't cut.");
             continue;
         }
         cl(cost);
         for(i=;i<n;i++)
             for(j=i+;j<n;j++)
                 cost[i][j]=cost[j][i]=dis(list[i],list[j]);
         for(i=;i<n;i++)
         {
             for(j=i;j<n;j++)dp[i][j]=INF;
             dp[i][(i+)%n]=;
         }
         for(i=n-;i>=;i--)
         {
             for(j=i+;j<n;j++)
             {
                 for(k=i+;k<=j-;k++)
                 {
                     dp[i][j]=min(dp[i][j],dp[i][k]+dp[k][j]+cost[i][k]+cost[k][j]);
                 }
             }
         }
         printf("%d\n",dp[][n-]);
     }
 }