zoukankan      html  css  js  c++  java
  • LightOJ 1057 位压缩dp

    /********************
    
    LightOJ 1057
    
    Author:Cdegree
    
    ********************/
    #include <cstdio>
    #include <cstdlib>
    #include <cstring>
    #include <cmath>
    #include <cctype>
    #include <vector>
    #include <stack>
    #include <queue>
    #include <map>
    #include <algorithm>
    #include <iostream>
    #include <string>
    #include <set>
    #define X first
    #define Y second
    #define sqr(x) (x)*(x)
    #pragma comment(linker,"/STACK:102400000,102400000")
    using namespace std;
    const double PI = acos(-1.0);
    map<int, int>::iterator it;
    typedef long long LL ;
    template<typename T> void checkmin(T &x, T y) {x = min(x, y);}
    template<typename T> void checkmax(T &x, T y) {x = max(x, y);}
    
    const int K = 17, N = 22;
    int dp[K][1<<K];
    char s[N][N];
    
    vector<pair<int, int> > pi;
    
    int dis[K][K];
    void pf(int mask, int n) {
        for(int i = 0; i < n; ++i) {
            if((mask >> i) & 1)putchar('1');
            else putchar('0');
        } puts("");
    }
    int main() {
        int T;
        scanf("%d", &T);
        for(int t = 1; t <= T; ++t) {
            int n, m;
            scanf("%d %d", &n, &m);
            for(int i = 0; i < n; ++i)scanf("%s", s[i]);
            pi.clear();
            for(int i = 0; i < n; ++i) {
                for(int j = 0; j < m; ++j) {
                    if(s[i][j] == 'x') {
                        pi.push_back(make_pair(i, j));
                    }
                }
            }
            for(int i = 0; i < n; ++i) {
                for(int j = 0; j < m; ++j) {
                    if(s[i][j] == 'g') {
                        pi.push_back(make_pair(i, j));
                    }
                }
            }
            n = pi.size();
            for(int i = 0; i < n; ++i) {
                for(int j = 0; j < n; ++j) {
                    dis[i][j] = max(abs(pi[i].X - pi[j].X) , abs(pi[i].Y - pi[j].Y));
                }
            }
            memset(dp, -1, sizeof(dp));
            dp[0][0] = 0;
            for(int mask = 0; mask < (1 << n); ++mask) {
                for(int i = 0; i < n; ++i) {
                    if(~dp[i][mask]) {
                        //pf(mask,n);
                        for(int j = 0; j < n; ++j) {
                            if(!((mask >> j) & 1)) {
                                if(~dp[j][mask|(1<<j)])checkmin(dp[j][mask|(1<<j)], dp[i][mask] + dis[i][j]);
                                else dp[j][mask|(1<<j)] = dp[i][mask] + dis[i][j];
                            }
                        }
                    }
                }
            }
            printf("Case %d: %d
    ", t, dp[0][(1<<n)-1]);
        }
        return 0;
    }
  • 相关阅读:
    Python中的浅复制、深复制
    Python
    CSS中
    Fluent_Python_Part3函数即对象,05-1class-func,一等函数,函数即对象
    Python
    本地简单HTTP服务器
    Fluent_Python_Part2数据结构,04-text-byte,文本和字节序列
    Fluent_Python_Part2数据结构,03-dict-set,字典和集合
    Fluent_Python_Part2数据结构,02-array-seq,序列类型
    Codeforces 246E Blood Cousins Return(树上启发式合并)
  • 原文地址:https://www.cnblogs.com/cxw199204/p/3349367.html
Copyright © 2011-2022 走看看