POJ 1178 Camelot（DP + 枚举）

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题意：8X8棋盘上有一个king和一些knight，问把他们集合到一个点，最小需要移动的步数，有一个条件，就是如果王遇到骑士，那么骑士就可以带着王装逼一起飞了

思路：先floyd求出整个期盼对于王和骑士的传递闭包，然后每次枚举一个终点，和一个王和某个骑士的相遇点，然后答案就是，某个骑士到相遇点在到终点最小值，然后不断维护最小值即可

代码：

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 64;const int d1[8][2] = {1, 2, 1, -2, 2, 1, 2, -1, -1, 2, -1, -2, -2, 1, -2, -1};const int d2[8][2] = {1, 1, 1, -1, 1, 0, 0, 1, 0, -1, -1, -1, -1, 0, -1, 1};const int INF = 0x3f3f3f3f;int g1[N][N], g2[N][N];char str[505];int king, knight[N], kn;int main() {    memset(g1, INF, sizeof(g1));    memset(g2, INF, sizeof(g2));    for (int i = 0; i < 64; i++) {g1[i][i] = 0;g2[i][i] = 0;    }    for (int i = 0; i < 8; i++) {for (int j = 0; j < 8; j++) {    for (int k = 0; k < 8; k++) {int x1 = i + d1[k][0];int y1 = j + d1[k][1];int x2 = i + d2[k][0];int y2 = j + d2[k][1];if (x1 >= 0 && x1 < 8 && y1 >= 0 && y1 < 8)    g1[i * 8 + j][x1 * 8 + y1] = 1;if (x2 >= 0 && x2 < 8 && y2 >= 0 && y2 < 8)    g2[i * 8 + j][x2 * 8 + y2] = 1;    }}    }    for (int k = 0; k < 64; k++) {for (int i = 0; i < 64; i++) {    for (int j = 0; j < 64; j++) {g1[i][j] = min(g1[i][j], g1[i][k] + g1[k][j]);g2[i][j] = min(g2[i][j], g2[i][k] + g2[k][j]);    }}    }    while (~scanf("%s", str)) {int len = strlen(str);king = (str[0] - 'A') * 8 + (str[1] - '0' - 1);kn = 0;for (int i = 2; i < len; i += 2)    knight[kn++] = (str[i] - 'A') * 8 + (str[i + 1] - '0' - 1);int ans = INF;for (int i = 0; i < 64; i++) {    for (int j = 0; j < 64; j++) {int sum = g2[king][j];for (int k = 0; k < kn; k++)    sum += g1[knight[k]][i];int tot = INF;for (int k = 0; k < kn; k++) {    if (sum - g1[knight[k]][i] + g1[knight[k]][j] + g1[j][i] < tot)tot = sum - g1[knight[k]][i] + g1[knight[k]][j] + g1[j][i];}ans = min(ans, tot);    }}printf("%d\n", ans);    }    return 0;}

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