1、問(wèn)題：只允許向下或者向右，求從左上到右下的最短距離，動(dòng)態(tài)規(guī)劃法

1	6	3	1	1	1
6	0	2	5	1	1
3	2	0	3	4	1
1	5	3	0	2	3
6	7	4	7	5	2
2	6	1	3	4	9

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1 public int minPathSum(int[][] grid) { 2 if(grid==null || grid.length==0) return 0; 3 int m = grid.length; 4 int n = grid[0].length; 5 int[][] p = new int[m][n]; 6 for (int i = 0; i < m; i++) { 7 for (int j = 0; j < n; j++) { 8 if (i == 0 && j == 0) 9 p[i][j] = grid[i][j]; 10 else if (i == 0 && j > 0) 11 p[i][j] = p[i][j-1] + grid[i][j]; 12 else if (i > 0 && j == 0) 13 p[i][j] = p[i-1][j] + grid[i][j]; 14 else 15 p[i][j] = Math.min(p[i-1][j], p[i][j-1]) + grid[i][j]; 16 } 17 } 18 return p[m-1][n-1]; 19 }

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2、問(wèn)題：允許上下左右走，求從左上到右下的最短距離

1	6	3	1	1	1
6	0	2	5	1	1
3	2	0	3	4	1
1	5	3	0	2	3
6	7	4	7	5	2
2	6	1	3	4	9

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每個(gè)方格看成是一個(gè)點(diǎn)，可以構(gòu)造N*M+2（源點(diǎn)和目標(biāo)點(diǎn)）個(gè)點(diǎn)的圖，然后用最短路徑算法求解

法1：下面是dijkstra算法

1 public void dijkstra(int[][] graph, int source){ 2 int n = graph.length; 3 int[] dis = new int[n]; 4 boolean[] set = new boolean[n]; 5 // 初始化 6 for(int i=0; i<n; i++){ 7 // -1表示不可達(dá) 8 dis[i] = graph[source][i]; 9 } 10 dis[source] = 0; 11 set[source] = true; 12 13 for(int i=0; i<n-1; i++){ 14 // 找到V中距離源點(diǎn)最近的點(diǎn) 15 int mindis = Integer.MAX_VALUE; 16 int minIndex=0; 17 for(int v=0; v<n; v++){ 18 if(!set[v] && dis[v]<mindis && dis[v]!=-1){ 19 mindis = dis[v]; 20 minIndex = v; 21 } 22 } 23 24 // 更新源點(diǎn)到V中每個(gè)點(diǎn)的距離 25 set[minIndex] = true; 26 for(int v=0; v<n; v++){ 27 if(!set[v] && graph[minIndex][v]>0 && dis[v] > dis[minIndex]+graph[minIndex][v]){ 28 dis[v] = dis[minIndex]+graph[minIndex][v]; 29 } 30 } 31 } 32 }

法2：用優(yōu)先隊(duì)列方法，廣度優(yōu)先搜索

優(yōu)先隊(duì)列存點(diǎn)v和源點(diǎn)到v的最短路徑。

代碼

1 package hihocoder; 2 3 import java.util.*; 4 5 public class WaterCity2{ 6 7 private PriorityQueue<Point> queue = new PriorityQueue<>(); 8 private Set<Integer> visited = new HashSet<>(); 9 private int n, m; 10 private int[] distN, distM; 11 private Point[][] street; 12 13 private class Point implements Comparable<Point>{ 14 int r; 15 int c; 16 int dis; 17 boolean blocked = false; 18 boolean visited = false; 19 20 private Point(int r, int c) { 21 this.r = r; 22 this.c = c; 23 } 24 25 private void visitNeighbors(){ 26 if(r!=0) update(street[r-1][c], distN[r-1]); 27 if(r!=n-1) update(street[r+1][c], distN[r]); 28 if(c!=0) update(street[r][c-1], distM[c-1]); 29 if(c!=m-1) update(street[r][c+1], distM[c]); 30 } 31 32 private void update(Point p, int len){ 33 if(!p.blocked && !p.visited && p.dis > dis + len){ 34 queue.remove(p); 35 p.dis = dis + len; 36 queue.add(p); 37 } 38 } 39 40 @Override 41 public int compareTo(Point p){ 42 return dis-p.dis; 43 } 44 } 45 46 private int minPath(int r0, int c0, int r1, int c1){ 47 Point start = street[r0][c0]; 48 Point end = street[r1][c1]; 49 queue.clear(); 50 visited.clear(); 51 for (int i = 0; i < n; i++) { 52 for (int j = 0; j < m; j++) { 53 street[i][j].visited = false; 54 street[i][j].dis = Integer.MAX_VALUE; 55 } 56 } 57 58 start.dis = 0; 59 queue.add(start); 60 while(!queue.isEmpty()){ 61 Point cur = queue.poll(); 62 if(cur == end) return cur.dis; 63 cur.visited = true; 64 cur.visitNeighbors(); 65 } 66 return -1; 67 } 68 69 public void deal() { 70 try(Scanner scanner = new Scanner(System.in)) { 71 while (scanner.hasNextInt()) { 72 n = scanner.nextInt(); 73 m = scanner.nextInt(); 74 distN = new int[n-1]; 75 distM = new int[m-1]; 76 for (int i = 0; i < n - 1; i++) { 77 distN[i] = scanner.nextInt(); 78 } 79 for (int i = 0; i < m - 1; i++) { 80 distM[i] = scanner.nextInt(); 81 } 82 83 street = new Point[n][m]; 84 for (int i = 0; i < n; i++) { 85 for (int j = 0; j < m; j++) { 86 street[i][j] = new Point(i, j); 87 } 88 } 89 90 int k = scanner.nextInt(); 91 for (int i = 0; i < k; i++) { 92 street[scanner.nextInt()-1][scanner.nextInt() - 1].blocked = true; 93 } 94 95 int q = scanner.nextInt(); 96 for (int i = 0; i < q; i++) { 97 System.out.println(minPath(scanner.nextInt()-1, scanner.nextInt()-1, 98 scanner.nextInt()-1, scanner.nextInt()-1)); 99 } 100 } 101 } 102 } 103 104 public static void main(String[] args){ 105 new WaterCity2().deal(); 106 } 107 }

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轉(zhuǎn)載于:https://www.cnblogs.com/shizhh/p/5728558.html

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