4281297_ac_0ms_232k.cpp

北大大牛代码 1240道题的原代码 超级权威

#include <vector>
#include <stdio.h>
#include <algorithm>

using namespace std;

int map[20][20];
int nv, np, mxd;
int st, ed;

void floyd()
{
	int i, j, k;

	for (i = 0; i < nv; i++)
	{
		for (j = 0; j < nv; j++)
		{
			if (map[j][i] && j != i)
			{
				for (k = 0; k < nv; k++)
				{
					if (map[i][k] && j != k && (map[j][k] == 0 || map[j][i] + map[i][k] < map[j][k]))
					{
						map[j][k] = map[j][i] + map[i][k];
					}
				}
			}
		}
	}
}

struct node
{
	int cost;
	vector <int> route;
};

vector <node> ans;
int visited[20];
int tmp[20][20];

bool cmp1(int a, int b)
{
	return visited[a] < visited[b];
}

void save(int cost)
{
	int i;
	node t;

	t.route.clear();
	t.cost = cost;
	for (i = 0; i < nv; i++)
	{
		if (visited[i])
		{
			t.route.push_back(i);
		}
	}
	sort(t.route.begin(), t.route.end(), cmp1);
	ans.push_back(t);
}

void dfs(int u, int cost, int depth)
{
	int i;

	visited[u] = depth;
	if (cost + map[u][ed] > mxd)
	{
		return ;
	}
	if (u == ed)
	{
		save(cost);
		return ;
	}
	for (i = 0; i < nv; i++)
	{
		if (!tmp[u][i] || visited[i])
		{
			continue;
		}
		dfs(i, cost + tmp[u][i], depth + 1);
		visited[i] = 0;
	}
}

bool cmp2(node a, node b)
{
	if (a.cost != b.cost)
	{
		return a.cost < b.cost;
	}
	int i = 0;

	while (i < a.route.size() && i < b.route.size() && a.route[i] == b.route[i])
		i++;
	if (a.route.empty())
	{
		return true;
	}
	if (b.route.empty())
	{
		return false;
	}
	return a.route[i] < b.route[i];
}

void solve()
{
	memset(visited, 0, sizeof visited);
	ans.clear();
	dfs(st, 0, 1);
	if (ans.empty())
	{
		puts(" NO ACCEPTABLE TOURS\n");
		return ;
	}
	sort(ans.begin(), ans.end(), cmp2);
	for (int i = 0; i < ans.size(); i++)
	{
		printf(" %d:", ans[i].cost);
		for (int j = 0; j < ans[i].route.size(); j++)
		{
			printf(" %d", ans[i].route[j] + 1);
		}
		puts("");
	}
	puts("");
}

int main()
{
	int i, c;
	int u, v;
	int cas = 1;

	while (true)
	{
		scanf("%d", &nv);
		if (nv == -1)
		{
			break;
		}
		printf("Case %d:\n", cas++);
		scanf("%d", &np);
		memset(map, 0, sizeof map);
		memset(tmp, 0, sizeof tmp);
		for (i = 0; i < np; i++)
		{
			scanf("%d%d%d", &u, &v, &c);
			u--;v--;
			map[u][v] = map[v][u] = c;
			tmp[u][v] = tmp[v][u] = c;
		}
		scanf("%d%d%d", &st, &ed, &mxd);
		st--;ed--;
		floyd();
		solve();
	}
	return 0;
}