hdu6582

Posted 2022-09-29 starve

题意：给定一个无向图，删除某些边有一定的代价，要求删掉使得最短路径减小，求最小代价。

分析：首先要spfa求出起点到各个点的最短距离。对于一条权值为w，起点为i，终点为j的边，设dis[k]为起点到k点的距离，若dis[j]=dis[i]+w，则将该边加入另一个图里，边的容量为删除这条边的代价，则从起点到终点的最大流即为答案。。

　　1、首先最短路径一定在最短路图上

　　2、如果起点和终点不联通，就不存在这样一条最短路径，所以最短路径一定会变大；

注意看范围。。wa17发

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<queue>
using namespace std;
typedef long long ll;
const ll INF=1e18;
const int M=1e4+4;
struct node
    ll u,v,nextt;
    ll w;
g[M<<2],e[M<<2];
ll s,t,tot1,tot2,cur[M],head1[M],head2[M],vis[M],deep[M];
ll dis[M];
void addedge1(ll u,ll v,ll w)
    g[tot1].u=u;
    g[tot1].v=v;
    g[tot1].w=w;
    g[tot1].nextt=head1[u];
    head1[u]=tot1++;

void addedge2(ll u,ll v,ll w)
    e[tot2].v=v;
    e[tot2].w=w;
    e[tot2].nextt=head2[u];
    head2[u]=tot2++;
    e[tot2].v=u;
    e[tot2].w=0;
    e[tot2].nextt=head2[v];
    head2[v]=tot2++;
    

void dij()
    for(int i=0;i<=t;i++)
        dis[i]=INF;
//    cout<<"!!"<<endl;
    queue<int>que;
    que.push(s);
    dis[s]=0;
    while(!que.empty())
        ll  u=que.front();
        que.pop();
        vis[u]=0;
        for(ll i=head1[u];~i;i=g[i].nextt)
            ll v=g[i].v;
            if(dis[v]>dis[u]+g[i].w)
                dis[v]=dis[u]+g[i].w;
                if(!vis[v])
                    vis[v]=1;
                    que.push(v);
                
            
        
    


ll dd[M];
bool bfs()
    memset(deep,0,sizeof(deep));
    queue<int>que;
    que.push(s);
    deep[s]=1;
    while(!que.empty())
        int u=que.front();
        que.pop();
        for(int i=head2[u];i!=-1;i=e[i].nextt)
            int v=e[i].v;
            if(e[i].w>0&&deep[v]==0)
                deep[v]=deep[u]+1;
                if(v==t)
                    return true;
                que.push(v);
            
        
    
    return deep[t]==0?false:true;

ll dfs(ll u,ll fl)
    if(u==t)
        return fl;
    ll ans=0,x=0;
    for(int i=cur[u];i!=-1;i=e[i].nextt)
        ll v=e[i].v;
        if(e[i].w>0&&deep[v]==deep[u]+1)
            x=dfs(v,min(e[i].w,fl-ans));
            e[i].w-=x;
            e[i^1].w+=x;
            if(e[i].w)
                cur[u]=i;
            ans+=x;
            if(ans==fl)
                return ans;
        
    
    if(ans==0)
        deep[u]=0;
    return ans;

ll dinic()
    ll ans=0;
    while(bfs())
        for(int i=0;i<=t;i++)
            cur[i]=head2[i];
        ans+=dfs(s,INF);
    
    return ans;

int main()
    ll test;
    scanf("%lld",&test);
    while(test--)
        ll n,m;
        tot1=tot2=0;
        scanf("%lld%lld",&n,&m);
    //    cout<<tot1<<"!!"<<tot2<<endl;
        s=1,t=n;
        
        for(int i=0;i<=n;i++)
            head1[i]=head2[i]=-1,vis[i]=0;
        while(m--)
            ll u,v;
            ll w;
            scanf("%lld%lld%lld",&u,&v,&w);
            addedge1(u,v,w);
        
        dij();
    //    for(int i=1;i<=n;i++)
    //        cout<<dis[i]<<endl;
         for (int i = 1; i <= n; i++)
             for (int j = head1[i]; ~j;j = g[j].nextt)
                 if (dis[g[j].u] + g[j].w == dis[g[j].v])
                    addedge2(g[j].u,g[j].v,g[j].w);
        printf("%lld\n",dinic());
    
    return 0;

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