• TSP问题


    Hie with the Pie

    题目传送:POJ - 3311 - Hie with the Pie

    AC代码:

    #include <map>
    #include <set>
    #include <list>
    #include <cmath>
    #include <deque>
    #include <queue>
    #include <stack>
    #include <bitset>
    #include <cctype>
    #include <cstdio>
    #include <string>
    #include <vector>
    #include <complex>
    #include <cstdlib>
    #include <cstring>
    #include <fstream>
    #include <sstream>
    #include <utility>
    #include <iostream>
    #include <algorithm>
    #include <functional>
    #define LL long long
    #define INF 0x7fffffff
    using namespace std;
    
    int n;
    
    int mp[15][15];
    int dp[1300][13];//dp[state][i]表示到达i点状态为state的最短距离
    
    void floyd() {
        for(int k = 0; k <= n; k ++) {
            for(int i = 0; i <= n; i ++) {
                for(int j = 0; j <= n; j ++) {
                    if(mp[i][j] > mp[i][k] + mp[k][j]) {
                        mp[i][j] = mp[i][k] + mp[k][j];
                    }
                }
            }
        }
    }
    
    int main() {
        while(scanf("%d", &n) != EOF) {
            if(n == 0) break;
            for(int i = 0; i <= n; i ++) {
                for(int j = 0; j <= n; j ++) {
                    scanf("%d", &mp[i][j]);
                }
            }
    
            floyd();
            memset(dp, 0x3f, sizeof(dp));
    
            for(int state = 0; state < (1 << n); state ++) {//枚举全部状态
                for(int i = 1; i <= n; i ++) {
                    if(state & (1 << (i - 1))) {//状态中已经经过了城市i
                        if(state == (1 << (i - 1))) {//状态中仅仅经过城市i
                            dp[state][i] = mp[0][i];
                        }
                        else {//状态中除了经过了城市i还经过了其它城市
                            for(int j = 1; j <= n; j ++) {
                                if(i != j && state & (1 << (j - 1))) {
                                    dp[state][i] = min(dp[state][i], dp[state ^ (1 << (i - 1))][j] + mp[j][i]);
                                }
                            }
                        }
                    }
                }
            }
    
            int ans = INF;
            for(int i = 1; i <= n; i ++) {
                ans = min(ans, dp[(1 << n) - 1][i] + mp[i][0]);
            }
            printf("%d
    ", ans);
        }
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/clnchanpin/p/7366538.html
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