ACM ICPC 乌鲁木齐网络赛 J. Our Journey of Dalian Ends

Life is a journey, and the road we travel has twists and turns, which sometimes lead us to unexpected places and unexpected people.

Now our journey of Dalian ends. To be carefully considered are the following questions.

Next month in Xian, an essential lesson which we must be present had been scheduled.

But before the lesson, we need to attend a wedding in Shanghai.

We are not willing to pass through a city twice.

All available expressways between cities are known.

What we require is the shortest path, from Dalian to Xian, passing through Shanghai.

Here we go.

Input Format

There are several test cases.

The first line of input contains an integer tt which is the total number of test cases.

For each test case, the first line contains an integer m~(m\le 10000)m (m≤10000) which is the number of known expressways.

Each of the following mm lines describes an expressway which contains two string indicating the names of two cities and an integer indicating the length of the expressway.

The expressway connects two given cities and it is bidirectional.

Output Format

For eact test case, output the shortest path from Dalian to Xian, passing through Shanghai, or output -1−1 if it does not exist.

样例输入

3

2

Dalian Shanghai 3

Shanghai Xian 4

5

Dalian Shanghai 7

Shanghai Nanjing 1

Dalian Nanjing 3

Nanjing Xian 5

Shanghai Xian 8

3

Dalian Nanjing 6

Shanghai Nanjing 7

Nanjing Xian 8

样例输出

7

12
-1

每个城市拆成出点和入点，源点连西安和大连，汇点连上海，相当于求从西安到上海和从大连到上海最小距离之和，每个城市入点和出点之间连一条容量为1的边，但是注意，上海的容量必须是2，再根据给出的边，分别连接出点入点，存入相应花费，那么问题就可以转化成最小费用最大流了，如果流量不为2输出-1，否则输出最小花费。

#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<map>
#include<queue>
#include<string>
using namespace std;
#define ll long long
const ll maxm = 10005;
const ll INF = 1e18 + 7;
struct node
{
ll u, v, flow, cost, next;
}edge[maxm * 10];
map<string, ll>p;
ll cnt, s, t, n, m, sum, FLOW;
ll head[maxm * 10], dis[maxm * 10], pre[maxm * 10];
char a[maxm], b[maxm];
void init()
{
p.clear();
cnt = 0, s = 0, t = n * 5 + 1, sum = 0, FLOW = 0;
memset(head, -1, sizeof(head));
}
void add(ll u, ll v, ll flow, ll cost)
{
edge[cnt].u = u, edge[cnt].v = v;
edge[cnt].flow = flow, edge[cnt].cost = cost;
edge[cnt].next = head[u], head[u] = cnt++;
edge[cnt].u = v, edge[cnt].v = u;
edge[cnt].flow = 0, edge[cnt].cost = -cost;
edge[cnt].next = head[v], head[v] = cnt++;
}
ll bfs()
{
queue<ll>q;
for (ll i = 0;i <= t;i++) dis[i] = INF;
memset(pre, -1, sizeof(pre));
dis[s] = 0, q.push(s);
ll rev = 0;
while (!q.empty())
{
ll u = q.front();q.pop();
for (ll i = head[u];i != -1;i = edge[i].next)
{
ll v = edge[i].v;
if (dis[v] > dis[u] + edge[i].cost&&edge[i].flow)
{
dis[v] = dis[u] + edge[i].cost;
pre[v] = i, q.push(v);
}
}
}
if (dis[t] == INF) return 0;
return 1;
}
ll MCMF()
{
ll ans = 0, minflow;
while (bfs())
{
minflow = INF;
for (ll i = pre[t];i != -1;i = pre[edge[i].u])
minflow = min(minflow, edge[i].flow);
for (ll i = pre[t];i != -1;i = pre[edge[i].u])
edge[i].flow -= minflow, edge[i ^ 1].flow += minflow;
ans += dis[t] * minflow;
FLOW += minflow;
}
return ans;
}
int main()
{
ll i, j, k, T, c;
scanf("%lld", &T);
while (T--)
{
scanf("%lld", &n);
init();
ll nn = n * 2;
for (i = 1;i <= n;i++)
{
scanf("%s%s%lld", a, b, &c);
if (p[a] == 0)
{
p[a] = ++sum, k = 1;
if (strcmp(a, "Shanghai") == 0) k = 2;
add(p[a], p[a] + nn, k, 0);
}
if (p[b] == 0)
{
p[b] = ++sum, k = 1;
if (strcmp(b, "Shanghai") == 0) k = 2;
add(p[b], p[b] + nn, k, 0);
}
ll u = p[a], v = p[b];
add(u + nn, v, INF, c);
add(v + nn, u, INF, c);
}
ll u = p["Dalian"];
add(s, u, 1, 0);
u = p["Xian"];
add(s, u, 1, 0);
u = p["Shanghai"];
add(u + nn, t, 2, 0);
ll ans = MCMF();
if (FLOW == 2) printf("%lld\n", ans);
else printf("-1\n");
}
return 0;
}