Barricade HDU - 5889（最短路+最小割）

Barricade

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2227    Accepted Submission(s): 655

Problem Description

The empire is under attack again. The general of empire is planning to defend his castle. The land can be seen as N towns and M roads, and each road has the same length and connects two towns. The town numbered 1 is where general's castle is located, and the town numbered N is where the enemies are staying. The general supposes that the enemies would choose a shortest path. He knows his army is not ready to fight and he needs more time. Consequently he decides to put some barricades on some roads to slow down his enemies. Now, he asks you to find a way to set these barricades to make sure the enemies would meet at least one of them. Moreover, the barricade on the i-th road requires wi units of wood. Because of lacking resources, you need to use as less wood as possible.

 

Input

The first line of input contains an integer t, then t test cases follow.
For each test case, in the first line there are two integers N(N≤1000) and M(M≤10000).
The i-the line of the next M lines describes the i-th edge with three integers u,v and w where 0≤w≤1000 denoting an edge between u and v of barricade cost w.

 

Output

For each test cases, output the minimum wood cost.

 

Sample Input

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

 

Sample Output

4

 

Source

2016 ACM/ICPC Asia Regional Qingdao Online

 

注意审题

题中说的是敌人会走最短的路 所以我们把所有的最短路都拿出来 跑一边最大流即可

怎样把所有的最短路拿出来 跑一边最短路即可。。。因为跑完之后 起点和终点之间的所有的最短路 可以通过判断 d[e.v] == d[e.u] + 1 来进行建网络流的图

注意建网络流图的方式 不然会t

#include <iostream>
#include <cstdio>
#include <sstream>
#include <cstring>
#include <map>
#include <cctype>
#include <set>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#define rap(i, a, n) for(int i=a; i<=n; i++)
#define rep(i, a, n) for(int i=a; i<n; i++)
#define lap(i, a, n) for(int i=n; i>=a; i--)
#define lep(i, a, n) for(int i=n; i>a; i--)
#define rd(a) scanf("%d", &a)
#define rlld(a) scanf("%lld", &a)
#define rc(a) scanf("%c", &a)
#define rs(a) scanf("%s", a)
#define MOD 2018
#define LL long long
#define ULL unsigned long long
#define Pair pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define _  ios_base::sync_with_stdio(0),cin.tie(0)
//freopen("1.txt", "r", stdin);
using namespace std;
const int maxn = , INF = 0x7fffffff;
int head[maxn], head2[maxn], dis[maxn], d[maxn], vis[maxn], cur[maxn];
int cnt, cnt2;
int n, m, s, t;

struct node
{
    int u, v, w, c, next;
}Node[maxn<<];

void add_(int u, int v, int w, int c)
{
    Node[cnt].u = u;
    Node[cnt].v = v;
    Node[cnt].w = w;
    Node[cnt].c = c;
    Node[cnt].next = head[u];
    head[u] = cnt++;
}

void add(int u, int v, int w, int c)
{
    add_(u, v, w, c);
    add_(v, u, w, c);
}

void spfa()
{
    for(int i=; i<=n; i++) dis[i] = INF;
    mem(vis, );
    queue<int> Q;
    Q.push(s);
    vis[s] = ;
    dis[s] = ;
    while(!Q.empty())
    {
        int u = Q.front(); Q.pop();
        vis[u] = ;
        for(int i=head[u]; i!=-; i=Node[i].next)
        {
            node e = Node[i];
            if(dis[e.v] > dis[u] + e.w)
            {
                dis[e.v] = dis[u] + e.w;
                if(!vis[e.v])
                {
                    vis[e.v] = ;
                    Q.push(e.v);
                }
            }
        }
    }
}

struct edge
{
    int u, v, c, next;
}Edge[maxn<<];

void add_edge(int u, int v, int c)
{
    Edge[cnt2].u = u;
    Edge[cnt2].v = v;
    Edge[cnt2].c = c;
    Edge[cnt2].next = head2[u];
    head2[u] = cnt2++;
}

void add_Edge(int u, int v, int c)
{
    add_edge(u, v, c);
    add_edge(v, u, );
}

bool bfs()
{
    queue<int> Q;
    mem(d, );
    Q.push(s);
    d[s] = ;
    while(!Q.empty())
    {
        int u = Q.front(); Q.pop();
        for(int i=head2[u]; i!=-; i=Edge[i].next)
        {
            edge e = Edge[i];
            if(!d[e.v] && e.c > )
            {
                d[e.v] = d[e.u] + ;
                Q.push(e.v);
                if(e.v == t) return ;
            }
        }
    }
    return d[t] != ;
}

int dfs(int u, int cap)
{
    int ret = , V;
    if(u == t || cap == )
        return cap;
    for(int &i=cur[u]; i!=-; i=Edge[i].next)
    {
        edge e = Edge[i];
        if(d[e.v] == d[e.u] +  && e.c > )
        {
            int V = dfs(e.v, min(cap, e.c));
            Edge[i].c -= V;
            Edge[i^].c += V;
            ret += V;
            cap -= V;
            if(cap == ) break;
        }
    }
    if(cap > ) d[u] = -;
    return ret;
}

int dinic(int u)
{
    int ans = ;
    while(bfs())
    {
        memcpy(cur, head2, sizeof(head2));
        ans += dfs(u, INF);
    }
    return ans;
}

void build()
{
    for(int i=; i<=n; i++)
        for(int j=head[i]; j!=-; j=Node[j].next)
        {
            node e = Node[j];
            if(dis[e.v] == dis[e.u] + )
                add_Edge(e.u, e.v, e.c);
        }
}

int main()
{
    int T;
    rd(T);
    while(T--)
    {
        mem(head, -);
        mem(head2, -);
        cnt = cnt2 = ;
        rd(n); rd(m);
        rep(i, , m)
        {
            int u, v, c;
            scanf("%d%d%d", &u, &v, &c);
            add(u, v, , c);
        }
        s = , t = n;
        spfa();
        build();
        printf("%d\n", dinic(s));
    }
    return ;
}