LightOJ 1061 1061 - N Queen Again（状压dp）

题意:

给定一个8∗8棋盘有8个皇后，问最少移动多少步可以使得皇后不互相攻击，移动的时候可以移动8个方向，但是皇后不能互相跨越

分析:

如果暴力移动的话,然后判断终态是否合法,复杂度非常高,但是我们提前预处理出所有8皇后的解的方案,只有92种
然后跑带权匹配就可以了,这里用状压dp就可以
问题来了,权值如何计算呢,有一个转化,皇后不能互相跨越这个条件其实是没用的,可以等价为先移动阻挡的皇后,两种需要的步数是一样的
问题解决,复杂度为O(92∗82∗28)

代码:

////  Created by TaoSama on 2015-12-10//  Copyright (c) 2015 TaoSama. All rights reserved.////#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;#define pr(x) cout << #x << " = " << x << "  "#define prln(x) cout << #x << " = " << x << endlconst int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;typedef pair<int, int> Point;int vis[3][20]; //column, main, vice diagonalPoint goal[8], s[8];int ans, g[8][8], f[10][1 << 8];int get(int i, int j) {    if(s[i].first == goal[j].first && s[i].second == goal[j].second) return 0;    if(s[i].first == goal[j].first || s[i].second == goal[j].second) return 1;    if(s[i].first - s[i].second == goal[j].first - goal[j].second) return 1;    if(s[i].first + s[i].second == goal[j].first + goal[j].second) return 1;    return 2;}void gao() {    for(int i = 0; i < 8; ++i)        for(int j = 0; j < 8; ++j)            g[i][j] = get(i, j);    memset(f, 0x3f, sizeof f);    f[0][0] = 0;    for(int i = 0; i < 8; ++i) {        for(int j = 0; j < 1 << 8; ++j) {            if(f[i][j] == INF) continue;            for(int k = 0; k < 8; ++k) {                if(j >> k & 1) continue;                f[i + 1][j | 1 << k] = min(f[i + 1][j | 1 << k], f[i][j] + g[i][k]);            }        }    }    ans = min(ans, f[8][(1 << 8) - 1]);}void dfs(int row) {    if(row == 8) {        gao();        return;    }    for(int i = 0; i < 8; ++i) {        if(vis[0][i] || vis[1][row - i + 8] || vis[2][row + i]) continue;        goal[row] = Point(row, i);        vis[0][i] = vis[1][row - i + 8] = vis[2][row + i] = 1;        dfs(row + 1);        vis[0][i] = vis[1][row - i + 8] = vis[2][row + i] = 0;    }}int main() {#ifdef LOCAL    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    int t; scanf("%d", &t);    int kase = 0;    while(t--) {        int idx = 0;        for(int i = 0; i < 8; ++i) {            char buf[10]; scanf("%s", buf);            for(int j = 0; j < 8; ++j)                if(buf[j] == 'q') s[idx++] = Point(i, j);        }        ans = INF;        dfs(0);        printf("Case %d: %d\n", ++kase, ans);    }    return 0;}