UVA - 658 最短路

思路：通过前后两种状态建立一条边，利用Dijsktra就可以做了。

注意利用二进制优化。

AC代码

#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <utility>
#include <string>
#include <iostream>
#include <map>
#include <set>
#include <vector>
#include <queue>
#include <stack>
using namespace std;
#pragma comment(linker, "/STACK:1024000000,1024000000")
#define eps 1e-10
#define inf 0x3f3f3f3f
#define PI pair<int, int>
typedef long long LL;
const int maxn = (1 << 20) + 5, maxm = 100 + 5;
int n, m, d[maxn], vis[maxn];
char before[maxm][25], after[maxm][25];
int cost[maxm];
struct Node{
	int bug, dist;
	Node(){}
	Node(int bug, int dis):bug(bug),dist(dis){}
	bool operator < (const Node& p) const {
		return dist > p.dist;
	}
};

int Dijsk(int u) {
	memset(d, inf, sizeof(d));
	memset(vis, 0, sizeof(vis));
	priority_queue<Node>Q;
	Q.push(Node(u, 0));
	d[u] = 0;
	while(!Q.empty()) {
		Node p = Q.top(); Q.pop();
		int u = p.bug;
		if(u == 0) return d[u];
		if(vis[u]) continue;
		vis[u] = 1;
		for(int i = 0; i < m; ++i) {
			bool ok = true;
			for(int j = 0; j < n; ++j) {
				if(before[i][j] == '+' && !(u & (1 << j))) {ok = false; break;}
				if(before[i][j] == '-' && (u & (1 << j))) {ok = false; break;}
			}
			if(!ok) continue; //不能打补丁
			Node v = Node(u, p.dist + cost[i]);
			for(int j = 0; j < n; ++j) {
				if(after[i][j] == '-') v.bug &= ~(1 << j);
				if(after[i][j] == '+') v.bug |= (1 << j);
			}
			if(v.dist < d[v.bug] || d[v.bug] < 0) {
				d[v.bug] = v.dist;
				Q.push(v);
			}
		}
	}
	return -1;
}

int main() {
	int kase = 0;
	while(scanf("%d%d", &n, &m) == 2 && n && m) {
		for(int i = 0; i < m; ++i) {
			scanf("%d%s%s", &cost[i], before[i], after[i]);
		}
		int ans = Dijsk((1 << n) - 1);
		printf("Product %d\n", ++kase);
		if(ans == -1) printf("Bugs cannot be fixed.\n");
		else printf("Fastest sequence takes %d seconds.\n", ans);
		printf("\n");
	}
	return 0;
}

如有不当之处欢迎指出！