HDU-5889-Barricade

ACM模版

描述

题解

给定 n 个结点和 m 条路径，路径的长度均为 1，破坏每条路径都有对应的代价，然后会有怪兽从 m 往 1 结点进攻，但是这些智障怪兽只会从最短路进攻，求代价的最小割。

那么我们先进行最短路求得最短路径，然后将最短路径添加到网络流系统中，求得最大流，也就是最小割。酱紫结果就出来了哦！！！模版题~~~

代码

#include <iostream>#include <cstring>#include <vector>#include <queue>using namespace std;const int MAXN = 2005;const int INF = 0x3f3f3f3f;struct node{    int v, w;    node(int v_, int w_) : v(v_), w(w_) {};};int n, m, vs, vt;int s, t;int dis[MAXN];int vis[MAXN];vector<node> e[MAXN];void spfa(){    memset(vis, 0, sizeof(vis));    memset(dis, 0x3f, sizeof(dis));    dis[s] = 0;    queue<int> q;    q.push(s);    while (!q.empty())    {        int u = q.front();        q.pop();        vis[u] = 0;        for (int i = 0; i < e[u].size(); i++)        {            int v = e[u][i].v;            if (dis[v] > dis[u] + 1)            {                dis[v] = dis[u] + 1;                if (!vis[v])                {                    q.push(v);                }                vis[v] = 1;            }        }    }}struct edge{    int from, to, cap, flow;    edge(int u, int v, int c, int f) : from(u), to(v), cap(c), flow(f) {}};struct Dinic{    int s, t;    vector<edge> edges;     //  边数的两倍    vector<int> G[MAXN];    //  邻接表，G[i][j]表示结点i的第j条边在e数组中的序号    bool vis[MAXN];         //  BFS使用    int dis[MAXN];          //  从起点到i的距离    int cur[MAXN];          //  当前弧下标    void init()    {        for (int i = 0; i <= n + 1; i++)        {            G[i].clear();        }        edges.clear();    }    void addEdge(int from, int to, int cap)    {        edges.push_back(edge(from, to, cap, 0));        edges.push_back(edge(to, from, 0, 0));        int sz = (int)edges.size();        G[from].push_back(sz - 2);        G[to].push_back(sz - 1);    }    bool bfs()    {        memset(vis, 0, sizeof(vis));        queue<int> q;        q.push(s);        dis[s] = 0;        vis[s] = 1;        while (!q.empty())        {            int x = q.front();            q.pop();            for (int i = 0; i < G[x].size(); i++)            {                edge &e = edges[G[x][i]];                if (!vis[e.to] && e.cap > e.flow)                {                    vis[e.to] = 1;                    dis[e.to] = dis[x] + 1;                    q.push(e.to);                }            }        }        return vis[t];    }    int dfs(int x, int a)    {        if (x == t || a == 0)        {            return a;        }        int flow = 0, f = 0;        for (int &i = cur[x]; i < G[x].size(); i++)        {            edge &e = edges[G[x][i]];            if (dis[x] + 1 == dis[e.to] && (f = dfs(e.to, min(a, e.cap - e.flow))) > 0)            {                e.flow += f;                edges[G[x][i] ^ 1].flow -= f;                flow += f;                a -= f;                if (a == 0)                {                    break;                }            }        }        return flow;    }    int Maxflow(int s, int t)    {        this->s = s;        this->t = t;        int flow = 0;        while (bfs())        {            memset(cur, 0, sizeof(cur));            flow += dfs(s, INF);        }        return flow;    }} dc;int main(){    int T;    scanf("%d", &T);    while (T--)    {        scanf("%d%d", &n, &m);        for (int i = 0; i <= n; i++)        {            e[i].clear();        }        int u, v, w;        for (int i = 1; i <= m; i++)        {            scanf("%d%d%d", &u, &v, &w);            e[u].push_back(node(v, w));            e[v].push_back(node(u, w));        }        s = 1, t = n;        spfa();        dc.init();        for (int i = 1; i <= n; i++)        {            for (int j = 0; j < e[i].size(); j++)            {                if (dis[e[i][j].v] == dis[i] + 1)                {                    dc.addEdge(i, e[i][j].v, e[i][j].w);                }            }        }        printf("%d\n", dc.Maxflow(s, t));    }    return 0;}