POJ2135

题目链接：

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

---

题目大意：

现在有 N 个节点，有M条边，要从 1 走到 N 然后再回到 1 。要求走的边不能重复，求最短路径。

---

解题过程：

之前看了最小费用最大流然后一直没有做题，于是找了一个模板题来刷，对着板子敲上去居然一次AC，然后又改了下最短路的算法，AC。

---

题目分析：

算是一个隐含的最小费用最大流，设每条边的容量为1，花费为路径长度。那么所求的就是一个从起点 1 到终点 N 流量为 2 的流的最小费用流。

这里用的是最短增广路算法。

AC代码：

#include <cstdio>#include <vector>#include <queue>#include <cstring>using namespace std;const int MAX = 1123, INF = 0x3f3f3f3f;int N, M;struct Node{    int to, cap, cost, rev;    Node(int to, int cap, int cost, int rev):to(to), cap(cap), cost(cost), rev(rev){}};vector<Node> edge[MAX];int dist[MAX], vis[MAX], prevv[MAX], preve[MAX];void add_edge(int from, int to, int cap, int cost) {    edge[from].push_back(Node(to, cap, cost, edge[to].size()));    edge[to].push_back(Node(from, 0, -cost, edge[from].size()-1));}int min_cost_flow(int s, int t, int f) {    int rst = 0;    //循环到到达了f流量    while (f > 0) {        memset(dist, INF, sizeof(dist));        queue<int> q;        q.push(s);        vis[s] = 1;        dist[s] = 0;        while (!q.empty()) {            int u = q.front();            for (int i = 0; i < edge[u].size(); i++) {                Node& e = edge[u][i];                if (e.cap > 0 && dist[e.to] > dist[u] + e.cost) {                    dist[e.to] = dist[u] + e.cost;                    prevv[e.to] = u;                    preve[e.to] = i;                    if (!vis[e.to]) {                        q.push(e.to);                        vis[e.to] = 1;                    }                }            }            q.pop();            vis[u] = 0;        }        if (dist[t] == INF) return -1;        //最路径上最小流量        int d = f;        for (int u = t; u != s; u = prevv[u]) {            d = min(d, edge[prevv[u]][preve[u]].cap);        }        //剩余的流量        f -= d;        //计算费用        rst += d * dist[t];        //修改路上所经过的边的容量        for (int u = t; u != s; u = prevv[u]) {            Node& e = edge[prevv[u]][preve[u]];            e.cap -= d;            edge[u][e.rev].cap += d;        }    }    return rst;}int main() {    scanf("%d %d", &N, &M);    for (int i = 0; i < M; i++) {        int from, to, cost;        scanf("%d %d %d", &from, &to, &cost);        add_edge(from, to, 1, cost);        add_edge(to, from, 1, cost);    }    //求流量为2的最小费用    int ans = min_cost_flow(1, N, 2);    printf("%d\n", ans);}

居然可以AC的Dijstra代码：

#include <cstdio>#include <vector>#include <queue>#include <cstring>using namespace std;const int MAX = 1123, INF = 0x3f3f3f3f;int N, M;struct Node{    int to, cap, cost, rev;    Node(int to, int cap, int cost, int rev):to(to), cap(cap), cost(cost), rev(rev){}};vector<Node> edge[MAX];int dist[MAX], vis[MAX], prevv[MAX], preve[MAX];void add_edge(int from, int to, int cap, int cost) {    edge[from].push_back(Node(to, cap, cost, edge[to].size()));    edge[to].push_back(Node(from, 0, -cost, edge[from].size()-1));}int min_cost_flow(int s, int t, int f) {    int rst = 0;    while (f > 0) {        memset(dist, INF, sizeof(dist));        queue<int> q;        q.push(s);        vis[s] = 1;        dist[s] = 0;        while (!q.empty()) {            int u = q.front();            for (int i = 0; i < edge[u].size(); i++) {                Node& e = edge[u][i];                if (e.cap > 0 && dist[e.to] > dist[u] + e.cost) {                    dist[e.to] = dist[u] + e.cost;                    prevv[e.to] = u;                    preve[e.to] = i;                    if (!vis[e.to]) {                        q.push(e.to);                        vis[e.to] = 1;                    }                }            }            q.pop();            vis[u] = 0;        }        if (dist[t] == INF) return -1;        int d = f;        for (int u = t; u != s; u = prevv[u]) {            d = min(d, edge[prevv[u]][preve[u]].cap);        }        f -= d;        rst += d * dist[t];        for (int u = t; u != s; u = prevv[u]) {            Node& e = edge[prevv[u]][preve[u]];            e.cap -= d;            edge[u][e.rev].cap += d;        }    }    return rst;}int main() {    scanf("%d %d", &N, &M);    for (int i = 0; i < M; i++) {        int from, to, cost;        scanf("%d %d %d", &from, &to, &cost);        add_edge(from, to, 1, cost);        add_edge(to, from, 1, cost);    }    int ans = min_cost_flow(1, N, 2);    printf("%d\n", ans);}