LightOJ 1061 N Queen Again(状压DP)

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题意：给定8皇后位置，问最少多少步使得其互不攻击。

思路：状压dp。

#include <algorithm>#include <iostream>#include <sstream>#include <cstring>#include <cstdio>#include <vector>#include <string>#include <queue>#include <stack>#include <cmath>#include <set>#include <map>using namespace std;typedef long long LL;#define mem(a, n) memset(a, n, sizeof(a))#define ALL(v) v.begin(), v.end()#define si(a) scanf("%d", &a)#define sii(a, b) scanf("%d%d", &a, &b)#define siii(a, b, c) scanf("%d%d%d", &a, &b, &c)#define pb push_back#define eps 1e-8const int inf = 0x3f3f3f3f, N = 10 + 5, MOD = 1e9 + 7;int T, cas = 0;int n, m;struct point {    int x, y;} points[10];char s[N][N];int num[N], dp[N][300];int tot, ans;int Dis(int i, int j) {    int d1 = abs(points[j].x - i);    int d2 = abs(points[j].y - num[i]);    int ret = 0;    if(d1) ret ++; if(d2) ret ++;    if(!abs(d1 - d2) && d1 && d2) ret --;    return ret;}int make(int i, int j) {    if(dp[i][j] != -1) return dp[i][j];    int ret = inf;    for(int k = 0; k < 8; k ++)         if(j & (1 << k))             ret = min(ret, Dis(i - 1, k) + make(i - 1, j - (1 << k)));    return dp[i][j] = ret;}void work() {    mem(dp, -1), dp[0][0] = 0;    ans = min(ans, make(8, 255));}void dfs(int k) {    if(k == 8) {        work();        return;    }    for(int j = 0; j < 8; j ++) {        num[k] = j;        bool flag = true;        for(int i = 0; i < k; i ++) {            if((num[i] == num[k]) || (i + num[i] == k + num[k]) || (i - num[i] == k - num[k])) {                flag = false;                break;            }        }        if(flag) dfs(k + 1);    }}int main(){#ifdef LOCAL    freopen("/Users/apple/input.txt", "r", stdin);//freopen("/Users/apple/out.txt", "w", stdout);#endif    si(T);    while(T --) {        tot = 0;        ans = inf;        for(int i = 0; i < 8; i ++) scanf("%s", s[i]);        for(int i = 0; i < 8; i ++) for(int j = 0; j < 8; j ++) {            if(s[i][j] == 'q') {                points[tot].x = i,                points[tot].y = j;                tot ++;            }        }        dfs(0);        printf("Case %d: %d\n", ++ cas, ans);    }        return 0;}

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