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HDU 5889 Barricade (最短路+最小割)

题目链接：点我~~

题意：1000个点10000条边的无向图，敌人从n走一条最短路到1，在第i条路设置障碍的代价是wi，求最少的代价使得敌人至少会遇到一次障碍。

思路：先确定最短路，然后在最短路径上跑一个网络流，即可求出最小割。

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> PI;
typedef pair< PI, int> PII;
const double eps=1e-5;
const double pi=acos(-1.0);
const int mod=1e9+7;
const int INF=0x3f3f3f3f;
#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1
const int MAXN = 1000+100;
const int MAXM = 20100;
struct node
{
    int v;
    int cost;
    node(int _v=0,int _cost=0):v(_v),cost(_cost) {}
};
vector<node>E[MAXN];
void add(int u,int v,int w)
{
    E[u].push_back(node(v,w));
}
bool vis[MAXN];
int cnt[MAXN];
int dist[MAXN];
bool SPFA(int start,int n)
{
    memset(vis,false,sizeof(vis));
    for(int i=1; i<=n; i++)dist[i]=INF;
    vis[start]=true;
    dist[start]=0;
    queue<int>que;
    while(!que.empty())que.pop();
    que.push(start);
    memset(cnt,0,sizeof(cnt));
    cnt[start]=1;
    while(!que.empty())
    {
        int u=que.front();
        que.pop();
        vis[u]=false;
        for(int i=0; i<E[u].size(); i++)
        {
            int v=E[u][i].v;
            if(dist[v]>dist[u]+1)
            {
                dist[v]=dist[u]+1;
                if(!vis[v])
                {
                    vis[v]=true;
                    que.push(v);
                }
            }
        }
    }
    return true;
}

struct Edge
{
    int to,next,cap,flow;
} edge[MAXM];
int tol;
int head[MAXN];
int gap[MAXN],dep[MAXN],pre[MAXN],cur[MAXN];
void init()
{
    tol = 0;
    memset(head,-1,sizeof(head));
}
void addedge(int u,int v,int w,int rw=0)
{
    edge[tol].to = v;
    edge[tol].cap = w;
    edge[tol].next = head[u];
    edge[tol].flow = 0;
    head[u] = tol++;
    edge[tol].to = u;
    edge[tol].cap = rw;
    edge[tol].next = head[v];
    edge[tol].flow = 0;
    head[v]=tol++;
}

int sap(int start,int end,int N)
{
    memset(gap,0,sizeof(gap));
    memset(dep,0,sizeof(dep));
    memcpy(cur,head,sizeof(head));
    int u = start;
    pre[u] = -1;
    gap[0] = N;
    int ans = 0;
    while(dep[start] < N)
    {
        if(u == end)
        {
            int Min = INF;
            for(int i = pre[u]; i != -1; i = pre[edge[i^1].to])
                if(Min > edge[i].cap - edge[i].flow)
                    Min = edge[i].cap - edge[i].flow;
            for(int i = pre[u]; i != -1; i = pre[edge[i^1].to])
            {
                edge[i].flow += Min;
                edge[i^1].flow -= Min;
            }
            u = start;
            ans += Min;
            continue;
        }
        bool flag = false;
        int v;
        for(int i = cur[u]; i != -1; i = edge[i].next)
        {
            v = edge[i].to;
            if(edge[i].cap - edge[i].flow && dep[v]+1 == dep[u])
            {
                flag = true;
                cur[u] = pre[v] = i;
                break;
            }
        }
        if(flag)
        {
            u = v;
            continue;
        }
        int Min = N;
        for(int i = head[u]; i != -1; i = edge[i].next)
            if(edge[i].cap - edge[i].flow && dep[edge[i].to] < Min)
            {
                Min = dep[edge[i].to];
                cur[u] = i;
            }
        gap[dep[u]]--;

        if(!gap[dep[u]])return ans;
        dep[u] = Min+1;
        gap[dep[u]]++;
        if(u != start) u = edge[pre[u]^1].to;
    }
    return ans;
}


int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,m;
        init();
        for(int i=0; i<=n; ++i)
            E[i].clear();
        scanf("%d%d",&n,&m);
        int u,v,w;
        for(int i=0; i<m; ++i)
        {
            scanf("%d%d%d",&u,&v,&w);
            add(u,v,w);
            add(v,u,w);
        }
        SPFA(1,n);
        for(int i=1; i<=n; ++i)
        {
            for(int j=0; j<E[i].size(); ++j)
            {
                if(dist[E[i][j].v]==dist[i]+1)
                    addedge(i,E[i][j].v,E[i][j].cost);
            }
        }
        printf("%d\n",sap(1,n,n));
    }
    return 0;
}