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Description: （2） 主要算法的基本思想：
从题目上来分析我认为这是一个图的最短路径问题。因此决定用Dijkstra算法按路径长度递增的顺序逐步产生最短路径的方法：设置两个顶点的集合T和S，集合S中存放已找到的最短路径的顶点，集合T中存放当前还未找到的最短路径的顶点。初始状态时，集合S中只包含源点V0，然后不断从集合T中选取到顶点V0路径长度最短的顶点加入到集合S中，集合S中每加入一个新的顶点U，都要修改顶点V0到集合T中剩余顶点的最短路径长度值，集合T中各顶点新的最短路径长度值为原来的最短路径长度值与顶点U的最短路径长度只值中的较小的。此过程不断重复，直到集合T的顶点全部加入到集合S为止。
-(2) the principal on the basic idea : onto analysis from the topics I think this is a map of the shortest path problem. It decided to use Dijkstra algorithm by increasing the path length of the order gradually emerge from the shortest path approach : set up two vertices of the pool T and S, S storage pools have found the shortest path to the peak, Set T has not kept current to find the shortest path to the peak. Initial state, the pool S contain only source V0. and then from the pool of selected T vertex V0 shortest path length of the apex into pools S, S pool each add a new peak U, vertex V0 be changed to pool the remaining vertices T is the shortest path length value, Set T vertices new shortest path length of the original value of the shortest path length and value of the vertex U short
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Size: 23915 |
Author: 林渊 |
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Description: 求图的顶点连通度算法。方法就是先对源和汇做枚举，之后对每个枚举情况，把除去源汇两点的其余所有顶点看成是容量限制为1的点，求网络的最大流，就是此点对的顶点连通度，之后对枚举的所有点对找连通度最小的当成图的连通度。
带有顶点容量限制的最大流方法：将带容量限制的顶点u拆成两个点u 和u*，原顶点u的入边为u 的入边，原顶点u的出边为u*的出边，之后在u 和u*之间连接双向边，边的容量为顶点的容量限制。-plans for the vertex connectivity algorithms. First is the method of sources and sinks so enumerated, after the enumeration of each, two sources and sinks to remove all remaining vertices as capacity constraints to a point, seeking the largest network flow, this is the apex of connectivity, followed by enumeration of all points of connectivity for the smallest of the plan as connectivity. Vertex capacity constraints with the biggest flow method : with the capacity constraints of the vertex is split into two u u and u *, the original entry point u u edge of the income side, the original point out u * u edge of the edge out after the u and u * a two-way link between the edge, while the capacity for the top point of the capacity constraints.
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Size: 1515 |
Author: 吴地瓜 |
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Description: 输入有向图的顶点，又有优先级的要求，通过此程序可以得到一个合理的排序-importation plan to the apex, another priority of the request, through this procedure can be a reasonable sort of
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Size: 1309 |
Author: 王想红 |
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Description: 贝齐尔曲线算法演示程序
实验原理：贝齐尔曲线的形状是通过一组多边折线（也称为贝齐尔控制多边形）的各顶点唯一的定义出来的。
程序思路：先用链表把鼠标点击过的点记录下来，做为控制多边形。然后调用n次Bernstein多项式的基函数运算出曲线的点,最后用直线分段画出来（本程序分成了50段来画）
-Bezier curve algorithm Experimental Demonstration Program Principle : Bezier curve's shape through a multilateral dogleg Group (also known as the Bezier control polygon) of the apex only in the definition years. Procedures ideas : first use the mouse to click the chain put off point on record, as a control polygon. Then called n-Bernstein polynomial function calculate the curve, Finally, subparagraph drawn straight (this program consists of 50 to paint)
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Size: 39301 |
Author: Ray |
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Description: 这是一个根据顶点进行三角剖分的vc程序,有做相关方面研究的朋友可以参考一下-under the apex Triangulation vc procedures, doing research in the relevant friends reference
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Size: 51191 |
Author: jaba |
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Description: altera关于hardcopy的资料PPT格式的
第3部分
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Size: 751819 |
Author: hubin |
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Description: WINDOWS APEX通信软件-WINDOWS APEX communications software
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Size: 33500 |
Author: 曹义 |
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Description: 多边形常用算法模块 1. 判断多边形是否简单多边形2. 检查多边形顶点的凸凹性 3. 判断多边形是否凸多边形 4. 求多边形面积5. 判断多边形顶点的排列方向，方法一 6. 判断多边形顶点的排列方向，方法二 7. 射线法判断点是否在多边形内8. 判断点是否在凸多边形内-Does the polygon commonly used algorithm module 1. judgements polygons whether simple polygon. 2. inspections polygons apex convex-concave. 3. judgements polygons whether raised - polygon. 4. ask the polygon area. 5. judgements polygons apex the arrangement direction, the method 1. 6. judgements polygons apex arrangement direction, the method 2. 7. beams methods to sentence the break point whether. 8. sentences the break point in the polygon whether in raised - polygon
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Size: 2062 |
Author: 孤星赶月 |
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Description: (1)输入E条弧<j,k>,建立AOE-网的存储结构 (2)从源点v出发,令ve[0]=0,按拓扑排序求其余各项顶点的最早发生时间ve[i](1<=i<=n-1).如果得到的拓朴有序序列中顶点个数小于网中顶点数n,则说明网中存在环,不能求关键路径,算法终止 否则执行步骤(3)(3)从汇点v出发,令vl[n-1]=ve[n-1],按逆拓朴排序求其余各顶点的最迟发生时间vl[i](n-2>=i>=2). (4)根据各顶点的ve和vl值,求每条弧s的最早发生时间e(s)和最迟开始时间l(s).若某条弧满足条件e(s)=l(s),则为关键活动.-(1) E importation of Arc lt; J, kgt; Establish AOE - network storage structure (2) v starting point source, ve [0] = 0, by topological sorting point for the rest of the earliest timing ve [i] (1LT ; = ilt; = n-1). if the Topography vertex orderly sequence number is less than net n vertices, a statement that net presence in Central, not for Critical Path, algorithm implementation steps to terminate or (3) (3) from the Department of v starting point, Vl [n-1] = ve [n-1], by the inverse order for the remaining topology of the latest occurrence of peak time Vl [i] (n-inversion; = IGT; = 2). (4) According to the apex of ve and Vl value for each s arc of the earliest timing e (s) and the latest starting time of l (s). If any meet the conditions of the arc e (s) = l (s), was critical activities.
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Size: 1972 |
Author: 叶兆源 |
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Description: 一个复杂的工程通常可以分解成一组小任务的集合，完成这些小任务意味着整个工程的完成。例如，汽车装配工程可分解为以下任务：将底盘放上装配线，装轴，将座位装在底盘上，上漆，装刹车，装门等等。任务之间具有先后关系，例如在装轴之前必须先将底板放上装配线。任务的先后顺序可用有向图表示——称为顶点活动（ Activity On Vertex, AOV）网络。有向图的顶点代表任务，有向边(i, j) 表示先后关系：任务j 开始前任务i 必须完成。图1 - 4显示了六个任务的工程，边（ 1 , 4）表示任务1在任务4开始前完成，同样边（ 4 , 6）表示任务4在任务6开始前完成，边（1 , 4）与（4 , 6）合起来可知任务1在任务6开始前完成，即前后关系是传递的。由此可知，边（1 , 4）是多余的，因为边（1 , 3）和（3 , 4）已暗示了这种关系。-a complex project can usually be divided into a group of small tasks set to complete these tasks means that the completion of the whole project. For example, car assembly projects can be divided into the following tasks : to the chassis via the assembly line, armored axis of the seat is installed in the chassis, painting and installing the brakes, armored doors, etc.. Between tasks has, for example, installed in the shaft bottom before they must first be placed on an assembly line. The task order is available to the map -- known as the vertex (Activity On Vertex, AOV) network. A plan to the apex representative tasks, to the edge (i, j) has said : j task before i task must be completed. Figure 1 - 4 6 shows the mandate of the project, while (1, 4) a mandate in the mission completed before t
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Size: 911136 |
Author: sdg |
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Description: 此程序使用最小生成树Prim算法。用户输入图的输入文件名，程序读此文件，并调用MinSpanTree()为图建立最小生成树。输出包括树的总权及其顶点和边的列表。-This procedure use the Prim algorithm. The
user input the filename of the chart, the procedure reads this document and transfers the chart to Minimum SpannirngTree with MinSpanTree () function. Output the table including tree s total power and its apex and side.
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Size: 7134 |
Author: 小山 |
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Description: 国外游戏开发者杂志2001年第四期配套代码，包括Jeff Lander的模拟从顶点的投影的程序代码--Code of Game Developer Magazine in 2001 4th issue, it includes code which simulates Jeff Lander that projection from apex
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Size: 2048 |
Author: 站长 |
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Description: 国外游戏开发者杂志2001年第三期配套代码，包含Jeff Lander的动画的顶点投影的源代码--Code of game development magazine in 2001 the third issue, contains Jeff Lander`s apex projection code of animation
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Size: 165888 |
Author: 站长 |
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Description: 神经网络\原始对偶算法求解过指定顶点的最短路- The nerve network \ primitive antithesis algorithm solves has
assigned the apex most to short-circuit
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Size: 17408 |
Author: 吴子杰 |
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Description: WINDOWS APEX通信软件-WINDOWS APEX communications software
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Size: 32768 |
Author: 曹义 |
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Description: 多边形常用算法模块 1. 判断多边形是否简单多边形2. 检查多边形顶点的凸凹性 3. 判断多边形是否凸多边形 4. 求多边形面积5. 判断多边形顶点的排列方向，方法一 6. 判断多边形顶点的排列方向，方法二 7. 射线法判断点是否在多边形内8. 判断点是否在凸多边形内-Does the polygon commonly used algorithm module 1. judgements polygons whether simple polygon. 2. inspections polygons apex convex-concave. 3. judgements polygons whether raised- polygon. 4. ask the polygon area. 5. judgements polygons apex the arrangement direction, the method 1. 6. judgements polygons apex arrangement direction, the method 2. 7. beams methods to sentence the break point whether. 8. sentences the break point in the polygon whether in raised- polygon
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Size: 2048 |
Author: 孤星赶月 |
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Description: (1)输入E条弧<j,k>,建立AOE-网的存储结构 (2)从源点v出发,令ve[0]=0,按拓扑排序求其余各项顶点的最早发生时间ve[i](1<=i<=n-1).如果得到的拓朴有序序列中顶点个数小于网中顶点数n,则说明网中存在环,不能求关键路径,算法终止 否则执行步骤(3)(3)从汇点v出发,令vl[n-1]=ve[n-1],按逆拓朴排序求其余各顶点的最迟发生时间vl[i](n-2>=i>=2). (4)根据各顶点的ve和vl值,求每条弧s的最早发生时间e(s)和最迟开始时间l(s).若某条弧满足条件e(s)=l(s),则为关键活动.-(1) E importation of Arc lt; J, kgt; Establish AOE- network storage structure (2) v starting point source, ve [0] = 0, by topological sorting point for the rest of the earliest timing ve [i] (1LT ; = ilt; = n-1). if the Topography vertex orderly sequence number is less than net n vertices, a statement that net presence in Central, not for Critical Path, algorithm implementation steps to terminate or (3) (3) from the Department of v starting point, Vl [n-1] = ve [n-1], by the inverse order for the remaining topology of the latest occurrence of peak time Vl [i] (n-inversion; = IGT; = 2). (4) According to the apex of ve and Vl value for each s arc of the earliest timing e (s) and the latest starting time of l (s). If any meet the conditions of the arc e (s) = l (s), was critical activities.
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Size: 2048 |
Author: 叶兆源 |
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Description: 一个复杂的工程通常可以分解成一组小任务的集合，完成这些小任务意味着整个工程的完成。例如，汽车装配工程可分解为以下任务：将底盘放上装配线，装轴，将座位装在底盘上，上漆，装刹车，装门等等。任务之间具有先后关系，例如在装轴之前必须先将底板放上装配线。任务的先后顺序可用有向图表示——称为顶点活动（ Activity On Vertex, AOV）网络。有向图的顶点代表任务，有向边(i, j) 表示先后关系：任务j 开始前任务i 必须完成。图1 - 4显示了六个任务的工程，边（ 1 , 4）表示任务1在任务4开始前完成，同样边（ 4 , 6）表示任务4在任务6开始前完成，边（1 , 4）与（4 , 6）合起来可知任务1在任务6开始前完成，即前后关系是传递的。由此可知，边（1 , 4）是多余的，因为边（1 , 3）和（3 , 4）已暗示了这种关系。-a complex project can usually be divided into a group of small tasks set to complete these tasks means that the completion of the whole project. For example, car assembly projects can be divided into the following tasks : to the chassis via the assembly line, armored axis of the seat is installed in the chassis, painting and installing the brakes, armored doors, etc.. Between tasks has, for example, installed in the shaft bottom before they must first be placed on an assembly line. The task order is available to the map-- known as the vertex (Activity On Vertex, AOV) network. A plan to the apex representative tasks, to the edge (i, j) has said : j task before i task must be completed. Figure 1- 4 6 shows the mandate of the project, while (1, 4) a mandate in the mission completed before t
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Size: 911360 |
Author: sdg |
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Description: apex SRC programing by vb.net
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Size: 114688 |
Author: haklam |
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Description: skema rangkaian apex b500
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Size: 32768 |
Author: setiawan
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