poj 2490 Pimp My Ride_poj - 2490-优快云博客

本文介绍了一个典型的状态动态规划问题及其实现代码。该问题通过计算最小成本来寻找最优解，并使用了位运算来表示状态。文章包含完整的C++实现代码。

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很明显的状态dp。

/*
 * Author: stormdpzh
 * Created Time: 2012/8/19 15:21:37
 * File Name: poj_2490.cpp
 */
#include <iostream>
#include <cstdio>
#include <sstream>
#include <cstring>
#include <string>
#include <cmath>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <algorithm>
#include <functional>

#define sz(v) ((int)(v).size())
#define rep(i, n) for(int i = 0; i < n; i++)
#define repf(i, a, b) for(int i = a; i <= b; i++)
#define repd(i, a, b) for(int i = a; i >= b; i--)
#define out(n) printf("%d\n", n)
#define mset(a, b) memset(a, b, sizeof(a))
#define wh(n) while(1 == scanf("%d", &n))
#define whz(n) while(1 == scanf("%d", &n) && n != 0)
#define lint long long

using namespace std;

const int INF = 1 << 30;
const int MaxN = 16;
const int MaxM = (1 << 14) ^ 1;

int cost[MaxN][MaxN];
int f[MaxN][MaxM];
int n;
int t;

void gao()
{
    for(int i = 0; i <= n; i++) for(int j = 0; j <= (1 << n); j++)
        f[i][j] = INF;
    for(int i = 0; i < n; i++) f[1][1 << i] = cost[i][i];
    for(int i = 2; i <= n; i++) for(int j = 0; j < (1 << n); j++) {
        if(f[i - 1][j] == INF) continue;
        for(int k = 0; k < n; k++) {
            if(j & (1 << k)) continue;
            int id = j | (1 << k);
            int c = f[i - 1][j] + cost[k][k];
            for(int l = 0; l < n; l++) {
                if(j & (1 << l)) c += cost[k][l];
            }
            f[i][id] = min(f[i][id], c);
        }
    }
}

int main()
{
    scanf("%d", &t);
    int ans = 1;
    while(t--) {
        scanf("%d", &n);
        int total = 0;
        rep(i, n) rep(j, n) scanf("%d", &cost[i][j]);
        gao();
        printf("Scenario #%d:\n", ans++);
        printf("You have officially been pimped for only $%d\n", f[n][(1 << n) - 1]);
        puts("");
    }
    return 0;
}