poj1178

题意：一个国际象棋棋盘，有一个国王，和若干个骑士（马）。问要把他们都挪到一个格需要至少多少步。（国王如果和某骑士在同一个格则两者可以同时以骑士的方式移动且只记做一步）

分析：暴力枚举，跳马距离可先用floyd求。枚举终点，枚举国王上马地点，枚举国王要遇到的骑士。

View Code

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
usingnamespace std;

#define inf 0x3f3f3f3f
#define maxn 70

struct Point
{
int x, y;
} king, knight[maxn];

char st[256];
int dist[10][10][10][10];
int n;

void floyd()
{
for (int i =0; i <8; i++)
for (int j =0; j <8; j++)
for (int k =0; k <8; k++)
for (int l =0; l <8; l++)
if ((abs(i - k) ==1&& abs(j - l) ==2) || (abs(i - k)
==2&& abs(j - l) ==1))
dist[i][j][k][l] =1;
else
dist[i][j][k][l] = inf;
for (int i =0; i <8; i++)
for (int j =0; j <8; j++)
dist[i][j][i][j] =0;
for (int i =0; i <64; i++)
for (int j =0; j <64; j++)
for (int k =0; k <64; k++)
{
int x1, x2, x3, y1, y2, y3;
x1 = i /8;
x2 = j /8;
x3 = k /8;
y1 = i %8;
y2 = j %8;
y3 = k %8;
dist[x2][y2][x3][y3] = min(dist[x2][y2][x3][y3],
dist[x2][y2][x1][y1] + dist[x1][y1][x3][y3]);
}
}

void make()
{
king.x = st[0] -'A';
king.y = st[1] -'1';
for (int i =2; i < strlen(st); i +=2)
{
int d = i /2-1;
knight[d].x = st[i] -'A';
knight[d].y = st[i +1] -'1';
}
n = strlen(st) /2-1;
}

void work()
{
if (n ==0)
{
printf("0\n");
return;
}
int ans = inf;
for (int i =0; i <64; i++)
{
int x = i /8;
int y = i %8;
int sum =0;
for (int j =0; j < n; j++)
sum += dist[x][y][knight[j].x][knight[j].y];
for (int j =0; j <64; j++)
{
Point cross;
cross.x = j /8;
cross.y = j %8;
int temp1 = max(abs(king.x - cross.x), abs(king.y - cross.y));
int temp2 = inf;
for (int k =0; k < n; k++)
temp2 = min(temp2, dist[x][y][cross.x][cross.y]
+ dist[cross.x][cross.y][knight[k].x][knight[k].y]
- dist[x][y][knight[k].x][knight[k].y]);
ans = min(ans, temp1 + temp2 + sum);
}
}
printf("%d\n", ans);
}

int main()
{
//freopen("t.txt", "r", stdin);
scanf("%s", st);
floyd();
make();
work();
return0;
}

转载于:https://www.cnblogs.com/rainydays/archive/2011/07/09/2101893.html