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 << endlconst 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;}