Description

给出一张\(N\)个点\(M\)条边的带权无向图，结点编号从\(1\)到\(N\)，求从\(1\)到\(N\)再到\(1\)的最短路，每条边最多走一次。

Input

第一行给出两个整数\(N\)和\(M\)，表示结点数和边数。
接下来的\(M\)行每行给出三个整数\(u\)，\(v\)和\(w\)，表示一边的两个端点和边权。
\(1 \leqslant N \leqslant 1000\)，\(1 \leqslant M \leqslant 10000\)。

Output

输出一个整数表示最短路径的长度。

Sample Input

4 5
1 2 1
2 3 1
3 4 1
1 3 2
2 4 2

Sample Output

6

Solution

要求两条不重叠的最短路就不能用简单最短路算法了，先求最短路，删边再求最短路也不行，于是用最小费用最大流。对于一条权重为\(w\)的无向边，花费为\(w\)的有向边，因为每条边最多走一次，因此容量设为\(1\)，又因为只求两条路径，故添加额外源点\(0\)，与结点\(1\)建一条容量为\(2\)花费为\(0\)的边，从源点\(0\)到汇点\(N\)跑最小费用最大流即可。

Code

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
typedef long long ll;
const int INF = 0x3f3f3f3f;
const int N = 1e3 + 10;
const int M = 4e4 + 10;struct Edge
{int from, to, next, cap, cost;Edge() {}Edge(int from, int to, int next, int cap, int cost) : from(from), to(to), next(next), cap(cap), cost(cost) {}
} edge[M];
int head[N], tot;void add(int from, int to, int cap, int cost)
{edge[tot] = Edge(from, to, head[from], cap, cost);head[from] = tot++;edge[tot] = Edge(to, from, head[to], 0, -cost);head[to] = tot++;
}void init()
{memset(head, -1, sizeof(head));tot = 0;
}int dis[N], pre[N];
bool vis[N];
queue<int> q;
bool spfa(int s, int t)
{while (!q.empty()) q.pop();memset(dis, 0x3f, sizeof(dis));memset(vis, false, sizeof(vis));memset(pre, -1, sizeof(pre));q.push(s); vis[s] = true; dis[s] = 0;     while (!q.empty()) {int u = q.front(); q.pop(); vis[u] = false;for (int i = head[u]; i != -1; i = edge[i].next) if (edge[i].cap && dis[u] + edge[i].cost < dis[edge[i].to]) {int v = edge[i].to;dis[v] = dis[u] + edge[i].cost;pre[v] = i;if (!vis[v]) q.push(v), vis[v] = true;}}return dis[t] < INF;
}int mcmf(int s, int t)
{int mincost = 0, maxflow = 0;while (spfa(s, t)) {int flow = INF;for (int i = pre[t]; i != -1; i = pre[edge[i].from]) flow = min(flow, edge[i].cap);for (int i = pre[t]; i != -1; i = pre[edge[i].from]) {edge[i].cap -= flow;edge[i ^ 1].cap += flow;mincost += edge[i].cost * flow;}maxflow += flow;}return mincost;
}int main()
{int n, m;scanf("%d%d", &n, &m);init();while (m--){int u, v, c;scanf("%d%d%d", &u, &v, &c);add(u, v, 1, c);add(v, u, 1, c);}add(0, 1, 2, 0);int ans = mcmf(0, n);printf("%d\n", ans);return 0;
}

Link

http://poj.org/problem?id=2135