USACO 3.3.3

很恶心的搜索...纠结了两天.

本来是一个简单的最短路问题，直接BFS或者SPFA或者什么别的都可以，但是题目中增加了王.而王的行动方式有三种：

1.王在骑士的路上，和骑士到汇合点

2.王走到骑士的路上，和骑士到汇合点

3.王自己去汇合点

 
前两种情况可以合并成一种，f[i,j][x,y]+f[x,y][m,n]-f[i,j][m,n]+max(abs(king.x-x),abs(king.y-y))

枚举每个骑士的出发点[i,j]和汇合点[m,n]，以及王和骑士的交汇点[x,y]，计算王到交汇点[x,y]需要走的距离.-f[i,j][m,n]可以视为将被枚举骑士走的路线重新计算(显然骑士原来走的距离是[i,j][m,n]).

 
第三种情况，王自己走过去的话就直接max(abs(king.x-m),abs(king.y-n))，加上总距离，取最小值即可.

一个比较重要的剪枝是，王和骑士的交汇点在王坐标±2的范围内，似乎是大牛们实验得出的结果.

 
需要特别说明的是坐标，题目中的r表示行数，c表示列数，后面的字母表示所在列，字母表示所在行.

需要注意的是，此类题目必须先建立并分析好大致模型，然后编写，进而剪枝，采用迭代式开发.

BTW，USACO不支持函数名重载，因此自己写abs的话会和<stdlib.h>里的abs冲突，进而编译错误.

 

Executing...

   Test 1: TEST OK [0.011 secs, 6476 KB]

   Test 2: TEST OK [0.011 secs, 6476 KB]

   Test 3: TEST OK [0.011 secs, 6480 KB]

   Test 4: TEST OK [0.043 secs, 6476 KB]

   Test 5: TEST OK [0.205 secs, 6480 KB]

   Test 6: TEST OK [0.313 secs, 6476 KB]

   Test 7: TEST OK [0.011 secs, 6476 KB]

   Test 8: TEST OK [0.011 secs, 6480 KB]

   Test 9: TEST OK [0.119 secs, 6476 KB]

   Test 10: TEST OK [0.464 secs, 6476 KB]

   Test 11: TEST OK [0.022 secs, 6480 KB]

   Test 12: TEST OK [0.000 secs, 6480 KB]

   Test 13: TEST OK [0.022 secs, 6480 KB]

   Test 14: TEST OK [0.000 secs, 6476 KB]

   Test 15: TEST OK [0.011 secs, 6476 KB]

   Test 16: TEST OK [0.011 secs, 6476 KB]

   Test 17: TEST OK [0.022 secs, 6480 KB]

   Test 18: TEST OK [0.011 secs, 6476 KB]

   Test 19: TEST OK [0.011 secs, 6476 KB]

   Test 20: TEST OK [0.000 secs, 6476 KB]

All tests OK.

Your program ('camelot') produced all correct answers!  This is your

submission #7 for this problem.  Congratulations!

 

[Code]

 

 

#include<stdio.h> #include<ctype.h> #include<string.h> #include<limits.h> #include<stdlib.h> struct point {int x, y;}; int knights = 0, x_, y_, dist[32][28][32][28], dx[] = {-1, -2, -2, -1, 1, 2, 2, 1}, dy[] = {2, 1, -1, -2, -2, -1, 1, 2}; bool vis[32][28]; point king, knight[781]; struct queue { int f, r, k, n, q[100000]; queue() {memset(q, 0, sizeof(q)); n = k = f= r = 0;} void push(int x){q[k++] = x; r++; n++;} int pop(){n--; return q[f++];} bool empty(){return (f<r) ? 0 : 1;} }; bool check(int x, int y) { if (x < 1 || x > x_ || y < 1 || y > y_) return 0; return 1; } int max(int x, int y) {return (x>y) ? x : y;} void bfs(int x, int y) { if (!check(x, y)) return; queue q; int u = (x-1)*y_ + y, _x = x, _y = y; vis[x][y] = true; dist[_x][_y][x][y] = 0; q.push(u); while(!q.empty()) { u = q.pop(); x = (u-1)/y_+1; y = (u-1)%y_+1; for (int d = 0; d < 8; d++) { int nx = x+dx[d], ny = y+dy[d]; if (check(nx, ny) && !vis[nx][ny]) { int v = (nx-1)*y_+ny; q.push(v); vis[nx][ny] = true; dist[_x][_y][nx][ny] = dist[nx][ny][_x][_y] = dist[_x][_y][x][y]+1; } } } } int main() { FILE *fin, *fout; fin = fopen("camelot.in", "r"); fout = fopen("camelot.out", "w"); int i, j, tmp = 0, s, k, ans = INT_MAX, i_, j_, ans_; char ch = 0; fscanf(fin, "%d%d", &x_, &y_); while (!isalpha(ch = getc(fin))); fscanf(fin, "%d", &tmp); king.y = ch-'A'+1; king.x = tmp; for (;;) { while (!isalpha(ch = getc(fin)) && !feof(fin)); if (feof(fin)) break; fscanf(fin, "%d", &tmp); knight[knights].y = ch-'A'+1; knight[knights++].x = tmp; } for (i = 1; i <= x_; i++) for (j = 1; j <= y_; j++) for (i_ = 1; i_ <= x_; i_++) for (j_ = 1; j_ <= y_; j_++) dist[i][j][i_][j_] = 1000000; for (i = 1; i <= x_; i++) for (j = 1; j <= y_; j++) { memset(vis, 0, sizeof(vis)); dist[i][j][i][j] = 0; bfs(i, j); } for (i = 1; i <= x_; i++) for (j = 1; j <= y_; j++) { ans_ = INT_MAX; s = 0; for (k = 0; k < knights; k++) { for (i_ = king.x-2; i_ <= king.x+2; i_++) for (j_ = king.y-2; j_ <= king.y+2; j_++) { if (!check(i_,j_)) continue; if (ans_ > dist[knight[k].x][knight[k].y][i_][j_]+dist[i_][j_][i][j]+ max(abs(i_-king.x),abs(j_-king.y))-dist[knight[k].x][knight[k].y][i][j]) ans_ = dist[i_][j_][i][j]+dist[knight[k].x][knight[k].y][i_][j_]+ max(abs(i_-king.x),abs(j_-king.y))-dist[knight[k].x][knight[k].y][i][j]; } s += dist[knight[k].x][knight[k].y][i][j]; } if (s+ans_<ans) ans = s+ans_; ans_ = max(abs(king.x-i),abs(king.y-j)); if (s+ans_<ans) ans = s+ans_; } fprintf(fout, "%d/n", ans); }