原題網(wǎng)址：http://bailian.openjudge.cn/practice/1729

思路：

　　方法1: 用點對表示兩個人的狀態(tài)，放在隊列中（隊列也可以用優(yōu)先隊列），當?shù)竭_一個點對的路徑上的最近距離大于先前求得的最近距離則進行更新。

注意：可能兩人的最近距離是0，在更新的時候要注意。

? ? ? 方法2: achievalbe(d2) 表示兩人在最小距離平方>= d2的情況下是否能夠到學校. 用距離的平方剛好能夠用二分來確定兩人能夠到學校的路線的最遠的距離.見代碼三.

詳細代碼：

普通隊列：

1 #include <cstdio> 2 #include <cstring> 3 #include <queue> 4 #include <algorithm> 5 #include <cmath> 6 #include <stack> 7 using namespace std; 8 9 class PointPair{ 10 public: 11 int r[2],c[2]; 12 PointPair(int r1=0, int c1=0, int r2=0, int c2=0){ 13 r[0]=r1, r[1]=r2; 14 c[0]=c1, c[1]=c2; 15 } 16 }home,tp; 17 int dis(int r1, int c1, int r2, int c2){ 18 return (r1-r2)*(r1-r2)+(c1-c2)*(c1-c2); 19 } 20 int dis(PointPair &a){ 21 return dis(a.r[0], a.c[0], a.r[1], a.c[1]); 22 } 23 24 queue<PointPair> que; 25 int n, dp[30][30][30][30][3], dx[5]={-1,1,0,0,0},dy[5]={0,0,-1,1,0}, 26 hr[2],hc[2],sr[2],sc[2]; 27 char mp[40][40],dir[10]="NSWE";// 28 29 void solve(){ 30 printf("%.2f\n", sqrt(dp[sr[0]][sc[0]][sr[1]][sc[1]][0])); 31 int d1 = dp[sr[0]][sc[0]][sr[1]][sc[1]][1], d2 = dp[sr[0]][sc[0]][sr[1]][sc[1]][2], r1 = sr[0], c1=sc[0], r2=sr[1], c2=sc[1]; 32 stack<int> stk[2]; 33 while(d1!=-1 && d2 !=-1){ 34 stk[0].push(d1), stk[1].push(d2); 35 r1 -= dx[d1], c1 -= dy[d1], 36 r2 -= dx[d2], c2 -= dy[d2]; 37 d1 = dp[r1][c1][r2][c2][1], d2 = dp[r1][c1][r2][c2][2]; 38 } 39 for(int i=0; i<2; ++i){ 40 while(!stk[i].empty() && stk[i].top()!=4){ 41 printf("%c", dir[stk[i].top()]); 42 stk[i].pop(); 43 } 44 printf("\n"); 45 } 46 printf("\n"); 47 } 48 49 int main(){ 50 while(scanf("%d", &n), n){ 51 memset(dp, -1, sizeof(dp)); 52 for(int i=0; i<n; ++i){ 53 scanf("%s", mp[i]); 54 for(int j=0; j<n; ++j){ 55 if(mp[i][j]=='H') 56 hr[0]=i, hc[0]=j; 57 else if(mp[i][j]=='h') 58 hr[1]=i, hc[1]=j; 59 else if(mp[i][j]=='S') 60 sr[0]=i, sc[0]=j; 61 else if(mp[i][j]=='s') 62 sr[1]=i,sc[1]=j; 63 } 64 } 65 home = PointPair(hr[0], hc[0], hr[1], hc[1]); 66 dp[hr[0]][hc[0]][hr[1]][hc[1]][0] = dis(home); 67 que.push(home); 68 while(!que.empty()){ 69 tp = que.front(); que.pop(); 70 for(int i=0; i<4; ++i){ 71 for(int j=0; j<4; ++j){ 72 int r1 = tp.r[0]+dx[i], c1 = tp.c[0]+dy[i], 73 r2 = tp.r[1]+dx[j], c2 = tp.c[1]+dy[j], 74 d1 = i, d2 = j; 75 if(r1<0 || c1<0 || r1==n || c1==n || 76 r2<0 || c2<0 || r2==n || c2==n || 77 mp[r1][c1]=='*' || mp[r2][c2]=='*' || 78 mp[r1][c1]=='s' || mp[r1][c1]=='h'|| 79 mp[r2][c2]=='S' || mp[r2][c2]=='H') 80 continue; 81 // 兩人全在school的狀態(tài)不會出現(xiàn)在隊列中 82 if(mp[tp.r[0]][tp.c[0]]=='S') 83 r1=tp.r[0], c1=tp.c[0], d1=4; 84 else if(mp[tp.r[1]][tp.c[1]]=='s') 85 r2=tp.r[1], c2=tp.c[1], d2=4; 86 int d = min(dis(r1,c1,r2,c2), dp[tp.r[0]][tp.c[0]][tp.r[1]][tp.c[1]][0]); 87 if(d <= dp[r1][c1][r2][c2][0]) continue;// 可添加最優(yōu)性剪枝 88 dp[r1][c1][r2][c2][0] = d; 89 dp[r1][c1][r2][c2][1] = d1; 90 dp[r1][c1][r2][c2][2] = d2; 91 if(mp[r1][c1]=='S' && mp[r2][c2]=='s') continue; 92 que.push(PointPair(r1,c1,r2,c2)); 93 } 94 } 95 } 96 solve(); 97 } 98 return 0; 99 }

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優(yōu)先隊列：

1 #include <cstdio> 2 #include <cstring> 3 #include <queue> 4 #include <algorithm> 5 #include <cmath> 6 #include <stack> 7 using namespace std; 8 9 int dis(int r1, int c1, int r2, int c2){ 10 return (r1-r2)*(r1-r2)+(c1-c2)*(c1-c2); 11 } 12 class PointPair{ 13 public: 14 int r[2],c[2]; 15 PointPair(int r1=0, int c1=0, int r2=0, int c2=0){ 16 r[0]=r1, r[1]=r2; 17 c[0]=c1, c[1]=c2; 18 } 19 int operator<(const PointPair &b) const{ 20 return dis(r[0],c[0],r[1],c[1])<dis(b.r[0],b.c[0],b.r[1],b.c[1]); 21 } 22 }home,tp; 23 int dis(PointPair &a){ 24 return dis(a.r[0], a.c[0], a.r[1], a.c[1]); 25 } 26 27 priority_queue<PointPair> que; 28 int n, dp[30][30][30][30][3], dx[5]={-1,1,0,0,0},dy[5]={0,0,-1,1,0}, 29 hr[2],hc[2],sr[2],sc[2], rev[5]={1,0,3,2,4}; 30 char mp[40][40],dir[10]="NSWE";// 31 32 void solve(){ 33 printf("%.2f\n", sqrt(dp[sr[0]][sc[0]][sr[1]][sc[1]][0])); 34 int d1 = dp[sr[0]][sc[0]][sr[1]][sc[1]][1], d2 = dp[sr[0]][sc[0]][sr[1]][sc[1]][2], r1 = sr[0], c1=sc[0], r2=sr[1], c2=sc[1]; 35 stack<int> stk[2]; 36 while(d1!=-1 && d2 !=-1){ 37 stk[0].push(d1), stk[1].push(d2); 38 r1 += dx[rev[d1]], c1 += dy[rev[d1]], 39 r2 += dx[rev[d2]], c2 += dy[rev[d2]]; 40 d1 = dp[r1][c1][r2][c2][1], d2 = dp[r1][c1][r2][c2][2]; 41 } 42 for(int i=0; i<2; ++i){ 43 while(!stk[i].empty() && stk[i].top()!=4){ 44 printf("%c", dir[stk[i].top()]); 45 stk[i].pop(); 46 } 47 printf("\n"); 48 } 49 printf("\n"); 50 } 51 52 int main(){ 53 while(scanf("%d", &n), n){ 54 memset(dp, -1, sizeof(dp)); 55 for(int i=0; i<n; ++i){ 56 scanf("%s", mp[i]); 57 for(int j=0; j<n; ++j){ 58 if(mp[i][j]=='H') 59 hr[0]=i, hc[0]=j; 60 else if(mp[i][j]=='h') 61 hr[1]=i, hc[1]=j; 62 else if(mp[i][j]=='S') 63 sr[0]=i, sc[0]=j; 64 else if(mp[i][j]=='s') 65 sr[1]=i,sc[1]=j; 66 } 67 } 68 home = PointPair(hr[0], hc[0], hr[1], hc[1]); 69 dp[hr[0]][hc[0]][hr[1]][hc[1]][0] = dis(home); 70 que.push(home); 71 while(!que.empty()){ 72 // tp = que.front(); que.pop(); 73 tp = que.top(); que.pop(); 74 for(int i=0; i<4; ++i){ 75 for(int j=0; j<4; ++j){ 76 int r1 = tp.r[0]+dx[i], c1 = tp.c[0]+dy[i], 77 r2 = tp.r[1]+dx[j], c2 = tp.c[1]+dy[j], 78 d1 = i, d2 = j; 79 if(r1<0 || c1<0 || r1==n || c1==n || 80 r2<0 || c2<0 || r2==n || c2==n || 81 mp[r1][c1]=='*' || mp[r2][c2]=='*' || 82 mp[r1][c1]=='s' || mp[r1][c1]=='h'|| 83 mp[r2][c2]=='S' || mp[r2][c2]=='H') 84 continue; 85 // 兩人全在school的狀態(tài)不會出現(xiàn)在隊列中 86 if(mp[tp.r[0]][tp.c[0]]=='S') 87 r1=tp.r[0], c1=tp.c[0], d1=4; 88 else if(mp[tp.r[1]][tp.c[1]]=='s') 89 r2=tp.r[1], c2=tp.c[1], d2=4; 90 int d = min(dis(r1,c1,r2,c2), dp[tp.r[0]][tp.c[0]][tp.r[1]][tp.c[1]][0]); 91 if(d <= dp[r1][c1][r2][c2][0]) continue;// 可添加最優(yōu)性剪枝 92 dp[r1][c1][r2][c2][0] = d; 93 dp[r1][c1][r2][c2][1] = d1; 94 dp[r1][c1][r2][c2][2] = d2; 95 if(mp[r1][c1]=='S' && mp[r2][c2]=='s') continue; 96 que.push(PointPair(r1,c1,r2,c2)); 97 } 98 } 99 } 100 solve(); 101 } 102 return 0; 103 } View Code

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方法二:

1 #include <cstdio> 2 #include <cmath> 3 #include <queue> 4 #include <stack> 5 #include <cstring> 6 #include <algorithm> 7 8 using namespace std; 9 10 char mp[40][40], dir[] = "NSWE", rdir[] = "SNEW"; 11 int n, dp[30][30][30][30], mk[30][30][30][30]; 12 int dy[4] = {0, 0, -1, 1}, dx[4] = {-1, 1, 0, 0}; 13 14 class Point{ 15 public: 16 int x, y; 17 Point(int _x=0, int _y=0): x(_x), y(_y) {} 18 }home[2], sch[2]; 19 20 21 int dis2(Point a, Point b){ 22 return (a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y); 23 } 24 25 26 bool not_in_map(int x){ 27 return x < 0 || x >= n; 28 } 29 30 31 int achievable(int d, int print=0){ 32 queue<int> que; 33 34 memset(dp, 0x8, sizeof(dp)); 35 36 if(min(dis2(home[0], home[1]), dis2(sch[0], sch[1])) < d) 37 return 0; 38 dp[home[0].x][home[0].y][home[1].x][home[1].y] = 0; 39 que.push(home[0].x); que.push(home[0].y); que.push(home[1].x); que.push(home[1].y); 40 41 while(!que.empty()){ 42 int x0, y0, x1, y1; 43 x0 = que.front(); que.pop(); y0 = que.front(); que.pop(); 44 x1 = que.front(); que.pop(); y1 = que.front(); que.pop(); 45 for(int i=0; i<4; ++i){ 46 for(int j=0; j<4; ++j){ 47 int tx0 = x0, ty0 = y0, tx1 = x1, ty1 = y1; 48 if(!(x0 == sch[0].x && y0 == sch[0].y)) 49 tx0 += dx[i], ty0 += dy[i]; 50 if(!(x1 == sch[1].x && y1 == sch[1].y)) 51 tx1 += dx[j], ty1 += dy[j]; 52 if(not_in_map(tx0) || not_in_map(ty0) || not_in_map(tx1) || not_in_map(ty1)) 53 continue; 54 if(mp[tx0][ty0] == '*' || mp[tx0][ty0] == 'S' || mp[tx0][ty0] == 'H') 55 continue; 56 if(mp[tx1][ty1] == '*' || mp[tx1][ty1] == 's' || mp[tx1][ty1] == 'h') 57 continue; 58 if(dis2(Point(tx0, ty0), Point(tx1, ty1)) < d) 59 continue; 60 if(dp[tx0][ty0][tx1][ty1] > dp[x0][y0][x1][y1] + 1){ 61 dp[tx0][ty0][tx1][ty1] = dp[x0][y0][x1][y1] + 1; 62 que.push(tx0); que.push(ty0); que.push(tx1); que.push(ty1); 63 if(mp[tx0][ty0] == 's' && mp[tx1][ty1] == 'S') 64 return 1; 65 } 66 } 67 } 68 } 69 70 return 0; 71 } 72 73 74 void print_path(){ 75 stack<char> p[2]; 76 int x0 = sch[0].x, y0 = sch[0].y, x1 = sch[1].x, y1 = sch[1].y; 77 78 while(!(mp[x0][y0] == 'h' && mp[x1][y1] == 'H')){ 79 for(int i=0; i<5; ++i){ 80 for(int j=0; j<5; ++j){ 81 int tx0 = x0, ty0 = y0, tx1 = x1, ty1 = y1; 82 if(i < 4 || mp[x0][y0] != 's') 83 tx0 += dx[i], ty0 += dy[i]; 84 if(j < 4 || mp[x1][y1] != 'S') 85 tx1 += dx[j], ty1 += dy[j]; 86 if(i == 4 && j == 4) 87 continue; 88 if(not_in_map(tx0) || not_in_map(ty0) || not_in_map(tx1) || not_in_map(ty1)) 89 continue; 90 if(!(dp[tx0][ty0][tx1][ty1] + 1 == dp[x0][y0][x1][y1])) 91 continue; 92 if(mp[tx0][ty0] != 's') 93 p[0].push(rdir[i]); 94 if(mp[tx1][ty1] != 'S') 95 p[1].push(rdir[j]); 96 if(mp[tx0][ty0] == 'h' && mp[tx1][ty1] == 'H'){ 97 for(int k=0; k<2; ++k){ 98 while(!p[1-k].empty()){ 99 printf("%c", p[1-k].top()); p[1-k].pop(); 100 } 101 printf("\n"); 102 } 103 return; 104 } 105 x0 = tx0, y0 = ty0, x1 = tx1, y1 = ty1; 106 goto out; 107 } 108 } 109 out: {} 110 } 111 } 112 113 114 int main(){ 115 freopen("input.txt", "r", stdin); 116 117 while(scanf("%d", &n) && n){ 118 for(int i=0; i<n; ++i){ 119 scanf("%s", mp[i]); 120 for(int j=0; j<n; ++j){ 121 switch(mp[i][j]){ 122 case 'h': home[0] = Point(i, j); break; 123 case 'H': home[1] = Point(i, j); break; 124 case 's': sch[0] = Point(i, j); break; 125 case 'S': sch[1] = Point(i, j); break; 126 default: break; 127 } 128 } 129 } 130 131 int mx = min(dis2(home[0], home[1]), dis2(sch[0], sch[1])) + 1, mn = 0; 132 while(mx > mn){ 133 int d = (mx + mn) >> 1; 134 if(achievable(d)) 135 mn = d + 1; 136 else 137 mx = d; 138 } 139 achievable(mx - 1); 140 printf("%.2f\n", sqrt(mx - 1)); 141 print_path(); 142 } 143 144 return 0; 145 } View Code

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轉載于:https://www.cnblogs.com/yyf2016/p/5789382.html

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