LightOJ 1057 - Collecting Gold （状压dp）

题意:

n∗m(n,m<20)的格子里有最多15个格子有金子,x是出发点,求出发并回来拿完金子的最短路程

分析:

一开始dp[x][y][s]:=在(x,y)金子状态为s的最短路程,然后空间炸了

然后就卡死了−−其实加上x包括15个金子看成tsp问题就好了,然后获得两个金子的路程,直接曼哈顿距离,因为是8个方向

状态dp[p][s]:=在哪个位置(x或者金子),x或者金子的状态为s的最短路程,然后就很裸了

代码:

//
//  Created by TaoSama on 2015-11-20
//  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 << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

int n, m, dp[16][1 << 16];
int g, x[16], y[16];
char s[25][25];

int getdis(int i, int j) {
return max(abs(x[i] - x[j]), abs(y[i] - y[j]));
}

int dfs(int p, int s) {
int& ret = dp[p][s];
if(~ret) return ret;
if(p == 0 && s == (1 << g) - 1) return 0;
ret = INF;
for(int i = 0; i < g; ++i) {
if(s >> i & 1) continue;
ret = min(ret, dfs(i, s | 1 << i) + getdis(p, i));
}
return ret;
}

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--) {
scanf("%d%d", &n, &m);
g = 1;
for(int i = 1; i <= n; ++i) {
scanf("%s", s[i] + 1);
for(int j = 1; j <= m; ++j) {
if(s[i][j] == 'x') x[0] = i, y[0] = j;
if(s[i][j] == 'g') x[g] = i, y[g++] = j;
}
}
memset(dp, -1, sizeof dp);
printf("Case %d: %d\n", ++kase, dfs(0, 0));
}
return 0;
}