P3959 [NOIP2017 提高组] 宝藏_p3959 [noip 2017 提高组] 宝藏模拟退火-优快云博客

本文链接：https://blog.youkuaiyun.com/Hardy_Lee/article/details/127035669

博客介绍了NOIP2017提高组比赛中的一道名为'宝藏'的问题，主要探讨了如何使用状态压缩动态规划进行求解。通过枚举子集并编写相应的C++代码，博主详细阐述了动态规划的思路和实现过程。

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①枚举j为i的非空子集：
for (int i = 1; i < (1 << n); ++i) for (int j = i; j; j = (j - 1) & i)

代码如下：

#include <bits/stdc++.h>

using namespace std;

inline int read() {
	int x = 0, f = 0; char ch = getchar();
	while (!isdigit(ch)) f = ch == '-', ch = getchar();
	while (isdigit(ch)) x = (x << 3) + (x << 1) + (ch ^ 48), ch = getchar();
	return f ? -x : x;
}

const int N = 12, inf = 0x3f3f3f3f;
int n, m, g[1 << N], d[N + 1][N + 1], f[1 << N][N]; 

int main() {
	n = read(); m = read(); 
	memset(d, 0x3f, sizeof d); 
	for (int i = 1; i <= n; ++i) d[i][i] = 0; 
	for (int i = 1; i <= m; ++i) {
		int u = read(), v = read(), w = read();
		d[u][v] = d[v][u] = min(d[u][v], w); 
	}
	for (int i = 1; i < (1 << n); ++i) {
		for (int j = 1; j <= n; ++j) {
			if ((i >> (j - 1)) & 1) {
				for (int k = 1; k <= n; ++k) {
					if (d[j][k] != inf) {
						g[i] |= (1 << (k - 1)); 
					}
				}
			}
		}
	}
	memset(f, 0x3f, sizeof f); 
	for (int i = 1; i <= n; ++i) f[1 << (i - 1)][0] = 0;  
	for (int i = 1; i < (1 << n); ++i) {
		for (int j = i; j; j = (j - 1) & i) { // 枚举i的子集 
			if ((g[j] & i) == i) {
				int remain = i ^ j, cost = 0; 
				for (int k = 1; k <= n; ++k) {
					if ((remain >> (k - 1)) & 1) {
						int t = inf; 
						for (int u = 1; u <= n; ++u) {
							if ((j >> (u - 1)) & 1) {
								t = min(d[u][k], t); 
							}
						}
						cost += t; 
					}
				}
				for (int k = 1; k < n; ++k) f[i][k] = min(f[i][k], f[j][k - 1] + cost * k); 
			}
		}
	}
	int ans = inf; 
	for (int i = 0; i < n; ++i) ans = min(ans, f[(1 << n) - 1][i]);  
	printf("%d\n", ans); 
	return 0;
}