A*搜寻算法俗称A星算法。A*算法是比较流行的启发式搜寻算法之一，被广泛套用于路径最佳化领域[。它的独特之处是检查最短路径中每个可能的节点时引入了全局信息，对当前节点距终点的距离做出估计，并作为评价该节点处于最短路线上的可能性的量度。
基本介绍
- 中文名：A*搜寻算法
- 外文名：A-star Algorithm
- 别名：A星算法
- 套用领域：图最优路径搜寻/查找
- 类型：求出最低通过成本的算法
算法描述
A*改变它自己行为的能力基于启发式代价函式，启发式函式在游戏中非常有用。在速度和精确度之间取得折衷将会让你的游戏运行得更快。在很多游戏中，你并不真正需要得到最好的路径，仅需要近似的就足够了。而你需要什幺则取决于游戏中发生着什幺，或者运行游戏的机器有多快。 假设你的游戏有两种地形，平原和山地，在平原中的移动代价是1而在山地的是3，那幺A星算法就会认为在平地上可以进行三倍于山地的距离进行等价搜寻。 这是因为有可能有一条沿着平原到山地的路径。把两个邻接点之间的评估距离设为1.5可以加速A*的搜寻过程。然后A*会将3和1.5比较，这并不比把3和1比较差。然而，在山地上行动有时可能会优于绕过山脚下进行行动。所以花费更多时间寻找一个绕过山的算法并不经常是可靠的。 同样的，想要达成这样的目标，你可以通过减少在山脚下的搜寻行为来打到提高A星算法的运行速率。若想如此可以将A星算法的山地行动耗费从3调整为2即可。这两种方法都会给出可靠地行动策略。
用C语言实现A*最短路径搜寻算法，作者 Tittup frog(跳跳蛙)。
#include <stdio.h>#include <math.h>#define MaxLength 100 //用于优先伫列（Open表）的数组#define Height 15 //地图高度#define Width 20 //地图宽度#define Reachable 0 //可以到达的结点#define Bar 1 //障碍物#define Pass 2 //需要走的步数#define Source 3 //起点#define Destination 4 //终点#define Sequential 0 //顺序遍历#define NoSolution 2 //无解决方案#define Infinity 0xfffffff#define East (1 << 0)#define South_East (1 << 1)#define South (1 << 2)#define South_West (1 << 3)#define West (1 << 4)#define North_West (1 << 5)#define North (1 << 6)#define North_East (1 << 7)typedef struct{ signed char x, y;} Point;const Point dir[8] ={ {0, 1}, // East {1, 1}, // South_East {1, 0}, // South {1, -1}, // South_West {0, -1}, // West {-1, -1}, // North_West {-1, 0}, // North {-1, 1} // North_East};unsigned char within(int x, int y){ return (x >= 0 && y >= 0 && x < Height && y < Width);}typedef struct{ int x, y; unsigned char reachable, sur, value;} MapNode;typedef struct Close{ MapNode *cur; char vis; struct Close *from; float F, G; int H;} Close;typedef struct //优先伫列（Open表）{ int length; //当前伫列的长度 Close* Array[MaxLength]; //评价结点的指针} Open;static MapNode graph[Height][Width];static int srcX, srcY, dstX, dstY; //起始点、终点static Close close[Height][Width];// 优先伫列基本操作void initOpen(Open *q) //优先伫列初始化{ q->length = 0; // 队内元素数初始为0}void push(Open *q, Close cls[Height][Width], int x, int y, float g){ //向优先伫列（Open表）中添加元素 Close *t; int i, mintag; cls[x][y].G = g; //所添加节点的坐标 cls[x][y].F = cls[x][y].G + cls[x][y].H; q->Array[q->length++] = &(cls[x][y]); mintag = q->length - 1; for (i = 0; i < q->length - 1; i++) { if (q->Array[i]->F < q->Array[mintag]->F) { mintag = i; } } t = q->Array[q->length - 1]; q->Array[q->length - 1] = q->Array[mintag]; q->Array[mintag] = t; //将评价函式值最小节点置于队头}Close* shift(Open *q){ return q->Array[--q->length];}// 地图初始化操作void initClose(Close cls[Height][Width], int sx, int sy, int dx, int dy){ // 地图Close表初始化配置 int i, j; for (i = 0; i < Height; i++) { for (j = 0; j < Width; j++) { cls[i][j].cur = &graph[i][j]; // Close表所指节点 cls[i][j].vis = !graph[i][j].reachable; // 是否被访问 cls[i][j].from = NULL; // 所来节点 cls[i][j].G = cls[i][j].F = 0; cls[i][j].H = abs(dx - i) + abs(dy - j); // 评价函式值 } } cls[sx][sy].F = cls[sx][sy].H; //起始点评价初始值 // cls[sy][sy].G = 0; //移步花费代价值 cls[dx][dy].G = Infinity;}void initGraph(const int map[Height][Width], int sx, int sy, int dx, int dy){ //地图发生变化时重新构造地 int i, j; srcX = sx; //起点X坐标 srcY = sy; //起点Y坐标 dstX = dx; //终点X坐标 dstY = dy; //终点Y坐标 for (i = 0; i < Height; i++) { for (j = 0; j < Width; j++) { graph[i][j].x = i; //地图坐标X graph[i][j].y = j; //地图坐标Y graph[i][j].value = map[i][j]; graph[i][j].reachable = (graph[i][j].value == Reachable); // 节点可到达性 graph[i][j].sur = 0; //邻接节点个数 if (!graph[i][j].reachable) { continue; } if (j > 0) { if (graph[i][j - 1].reachable) // left节点可以到达 { graph[i][j].sur |= West; graph[i][j - 1].sur |= East; } if (i > 0) { if (graph[i - 1][j - 1].reachable && graph[i - 1][j].reachable && graph[i][j - 1].reachable) // up-left节点可以到达 { graph[i][j].sur |= North_West; graph[i - 1][j - 1].sur |= South_East; } } } if (i > 0) { if (graph[i - 1][j].reachable) // up节点可以到达 { graph[i][j].sur |= North; graph[i - 1][j].sur |= South; } if (j < Width - 1) { if (graph[i - 1][j + 1].reachable && graph[i - 1][j].reachable && map[i][j + 1] == Reachable) // up-right节点可以到达 { graph[i][j].sur |= North_East; graph[i - 1][j + 1].sur |= South_West; } } } } }}int bfs(){ int times = 0; int i, curX, curY, surX, surY; unsigned char f = 0, r = 1; Close *p; Close* q[MaxLength] = { &close[srcX][srcY] }; initClose(close, srcX, srcY, dstX, dstY); close[srcX][srcY].vis = 1; while (r != f) { p = q[f]; f = (f + 1) % MaxLength; curX = p->cur->x; curY = p->cur->y; for (i = 0; i < 8; i++) { if (! (p->cur->sur & (1 << i))) { continue; } surX = curX + dir[i].x; surY = curY + dir[i].y; if (! close[surX][surY].vis) { close[surX][surY].from = p; close[surX][surY].vis = 1; close[surX][surY].G = p->G + 1; q[r] = &close[surX][surY]; r = (r + 1) % MaxLength; } } times++; } return times;}int astar(){ // A*算法遍历 //int times = 0; int i, curX, curY, surX, surY; float surG; Open q; //Open表 Close *p; initOpen(&q); initClose(close, srcX, srcY, dstX, dstY); close[srcX][srcY].vis = 1; push(&q, close, srcX, srcY, 0); while (q.length) { //times++; p = shift(&q); curX = p->cur->x; curY = p->cur->y; if (!p->H) { return Sequential; } for (i = 0; i < 8; i++) { if (! (p->cur->sur & (1 << i))) { continue; } surX = curX + dir[i].x; surY = curY + dir[i].y; if (!close[surX][surY].vis) { close[surX][surY].vis = 1; close[surX][surY].from = p; surG = p->G + sqrt((curX - surX) * (curX - surX) + (curY - surY) * (curY - surY)); push(&q, close, surX, surY, surG); } } } //printf("times: %d\n", times); return NoSolution; //无结果}const int map[Height][Width] = { {0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,0,1,1}, {0,0,1,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1}, {0,0,0,0,0,0,1,0,0,0,0,0,0,1,1,0,0,0,0,1}, {0,0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,0,0,0}, {0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0,1}, {0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0}, {0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0}, {0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0}, {0,0,0,1,0,0,0,0,0,1,1,0,0,0,0,0,0,0,0,0}, {0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0}, {0,1,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0}, {0,0,0,0,1,0,0,1,0,0,0,0,1,0,0,0,0,0,0,0}, {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,0}, {0,1,0,0,0,0,1,0,0,0,0,0,0,1,0,1,0,0,0,1}, {0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0}};const char Symbol[5][3] = { "□", "▓", "▽", "☆", "◎" };void printMap(){ int i, j; for (i = 0; i < Height; i++) { for (j = 0; j < Width; j++) { printf("%s", Symbol[graph[i][j].value]); } puts(""); } puts("");}Close* getShortest(){ // 获取最短路径 int result = astar(); Close *p, *t, *q = NULL; switch(result) { case Sequential: //顺序最近 p = &(close[dstX][dstY]); while (p) //转置路径 { t = p->from; p->from = q; q = p; p = t; } close[srcX][srcY].from = q->from; return &(close[srcX][srcY]); case NoSolution: return NULL; } return NULL;}static Close *start;static int shortestep;int printShortest(){ Close *p; int step = 0; p = getShortest(); start = p; if (!p) { return 0; } else { while (p->from) { graph[p->cur->x][p->cur->y].value = Pass; printf("（%d，%d）→\n", p->cur->x, p->cur->y); p = p->from; step++; } printf("（%d，%d）\n", p->cur->x, p->cur->y); graph[srcX][srcY].value = Source; graph[dstX][dstY].value = Destination; return step; }}void clearMap(){ // Clear Map Marks of Steps Close *p = start; while (p) { graph[p->cur->x][p->cur->y].value = Reachable; p = p->from; } graph[srcX][srcY].value = map[srcX][srcY]; graph[dstX][dstY].value = map[dstX][dstY];}void printDepth(){ int i, j; for (i = 0; i < Height; i++) { for (j = 0; j < Width; j++) { if (map[i][j]) { printf("%s ", Symbol[graph[i][j].value]); } else { printf("%2.0lf ", close[i][j].G); } } puts(""); } puts("");}void printSur(){ int i, j; for (i = 0; i < Height; i++) { for (j = 0; j < Width; j++) { printf("%02x ", graph[i][j].sur); } puts(""); } puts("");}void printH(){ int i, j; for (i = 0; i < Height; i++) { for (j = 0; j < Width; j++) { printf("%02d ", close[i][j].H); } puts(""); } puts("");}int main(int argc, const char **argv){ initGraph(map, 0, 0, 0, 0); printMap(); while (scanf("%d %d %d %d", &srcX, &srcY, &dstX, &dstY) != EOF) { if (within(srcX, srcY) && within(dstX, dstY)) { if (shortestep = printShortest()) { printf("从（%d，%d）到（%d，%d）的最短步数是: %d\n", srcX, srcY, dstX, dstY, shortestep); printMap(); clearMap(); bfs(); //printDepth(); puts((shortestep == close[dstX][dstY].G) ? "正确" : "错误"); clearMap(); } else { printf("从（%d，%d）不可到达（%d，%d）\n", srcX, srcY, dstX, dstY); } } else { puts("输入错误！"); } } return (0);}性质
可採纳性
在一些问题的求解过程中，如果存在最短路径，无论在什幺情况之下，如果一个搜寻算法都能够保证找到这条最短路径，则称这样的搜寻算法就具有可採纳性。
单调性
A*算法扩展的所有节点的序列的f值是递增的，那幺它最先生成的路径一定就是最短的。
信息性
如何判断两种策略哪一个更好呢?具有的启发信息越多，搜寻的状态就越少，越容易找到最短路径，这样的策略就更好。在两个启发策略中，如果有h(n1)
h(n2)，则表示策略h:比h,具有更多的启发信息，同时h(n)越大表示它所搜寻的空间数就越少。但是同时需要注意的是:信息越多就需要更多的计算时间，从而有可能抵消因为信息多而减少的搜寻空间所带来的好处。
缺陷
A*算法进行下一步将要走的节点的搜寻的时候，每次都是选择F值最小的节点，因此找到的是最优路径。但是正因为如此A*算法每次都要扩展当前节点的全部后继节点，运用启发函式计算它们的F值，然后选择F值最小的节点作为下一步走的节点。在这个过程中，OPEN表需要保存大量的节点信息，不仅存储量大是一个问题，而且在查找F值最小的节点时，需要查询的节点也非常多，当然就非常耗时，这个问题就非常严重了。再加上如果游戏地图庞大，路径比较複杂，路径搜寻过程则可能要计算成千上万的节点，计算量非常巨大。因此，搜寻一条路径需要一定的时间，这就意味着游戏运行速度降低。
