ZOJ 2027 Travelling Fee

大意不再赘述。

思路：最直接的思路是通过暴力枚举将每一条边置为0，求一次最短路，然后再恢复，正解应该是通过Floyd算法来求解的，但我没想出来。

CODE：

#include <iostream>
#include <cstdlib>
#include <cstring>
#include <string>
#include <cstdio>
#include <map>
#include <queue>
using namespace std;

const int MAXN = 110;
const int MAXM = 10010;
const int INF = 0x3f3f3f3f;

map<string, int> Map;

struct Edge
{
	int v, w;
	int next;
}edge[MAXM];

int first[MAXN], d[MAXN];
char str1[MAXN], str2[MAXN];
int n, m;
int cnt;
int tot;

void init()
{
	cnt = 0;
	tot = 2;
	memset(first, -1, sizeof(first));
	Map.clear();
}

void spfa(int src)
{
	queue<int> q;
	bool inq[MAXN] = {0};
	for(int i = 1; i <= tot; i++) d[i] = (i == src)?0:INF;
	q.push(src);
	while(!q.empty())
	{
		int x = q.front(); q.pop();
		inq[x] = 0;
		for(int e = first[x]; e != -1; e = edge[e].next)
		{
			int v = edge[e].v, w = edge[e].w;
			if(d[v] > d[x] + w)
			{
				d[v] = d[x] + w;
				if(!inq[v])
				{
					q.push(v);
					inq[v] = 1;
				}
			}
		}
	}
}

void read_graph(int u, int v, int w)
{
	edge[cnt].v = v, edge[cnt].w = w;
	edge[cnt].next = first[u], first[u] = cnt++;
}

void read_case()
{
	scanf("%d", &m);
	while(m--)
	{
		int u, v, w;
		scanf("%s%s%d", str1, str2, &w);
		if(!Map[str1]) Map[str1] = ++tot;
		if(!Map[str2]) Map[str2] = ++tot;
		u = Map[str1], v = Map[str2];
		read_graph(u, v, w);
		read_graph(v, u, w);
	}
}

void solve()
{
	int ans = INF;
	for(int u = 1; u <= tot; u++)
	{
		for(int e = first[u]; e != -1; e = edge[e].next)
		{
			int t = edge[e].w;
			edge[e].w = 0;
			spfa(1);
			ans = min(ans, d[2]);
			edge[e].w = t;
		}
	}
	printf("%d\n", ans);
}

int main()
{
	while(~scanf("%s%s", str1, str2))
	{
		init();
		Map[str1] = 1;
		Map[str2] = 2;
		read_case();
		solve();
	}
	return 0;
}