hdu 5294 Tricks Device最短路+最小割多校連合第1場

7675 ワード

Tricks Device
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 66 Accepted Submission(s): 14
Problem Description
Innocent Wu follows Dumb Zhang into a ancient tomb. Innocent Wu’s at the entrance of the tomb while Dumb Zhang’s at the end of it. The tomb is made up of many chambers, the total number is N. And there are M channels connecting the chambers. Innocent Wu wants to catch up Dumb Zhang to find out the answers of some questions, however, it’s Dumb Zhang’s intention to keep Innocent Wu in the dark, to do which he has to stop Innocent Wu from getting him. Only via the original shortest ways from the entrance to the end of the tomb costs the minimum time, and that’s the only chance Innocent Wu can catch Dumb Zhang.
Unfortunately,Dumb Zhang masters the art of becoming invisible(奇門遁甲)and tricks devices of this tomb, he can cut off the connections between chambers by using them. Dumb Zhang wanders how many channels at least he has to cut to stop Innocent Wu. And Innocent Wu wants to know after how many channels at most Dumb Zhang cut off Innocent Wu still has the chance to catch Dumb Zhang.

Input
There are multiple test cases. Please process till EOF.
For each case,the first line must includes two integers, N(<=2000), M(<=60000). N is the total number of the chambers, M is the total number of the channels.
In the following M lines, every line must includes three numbers, and use ai、bi、li as channel i connecting chamber ai and bi(1<=ai,bi<=n), it costs li(0The entrance of the tomb is at the chamber one, the end of tomb is at the chamber N.

Output
Output two numbers to stand for the answers of Dumb Zhang and Innocent Wu’s questions.

Sample Input

Sample Output

2 6

  図Nの点Mの辺がなくて、出力は少なくともいくつかの辺を削除して最短路を破壊して、最大でいくつかの辺を削除して最短路を破壊しません.
  最短路を破壊しない答えはMから最短路の中の数が最も少ないものを差し引いて、少なくともいくつか削除して最短路を破壊する方法は、最短路上のすべての辺を再構築して、流量が1のネットワークの流れになって、最小割を求めることです.
  最短ルートを求めるときに前のノードを記録します.

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<cmath>
#include<vector>
#include<queue>

using namespace std;
typedef long long LL;

const int INF=0x3f3f3f3f;
const int MAXN=2010;

int N,M;

struct Edge{
    int from,to,cap,flow;
};

struct Dinic{
    int n,m,s,t;            //   ，  （     ），         
    vector<Edge> edges;     //  。edge[e] edge[e^1]     
    vector<int> G[MAXN];    //   ，G[i][j]    i  j   e      
    bool vis[MAXN];         //BFS  
    int d[MAXN];            //    i   
    int cur[MAXN];          //     

    void clear_all(int n){
        for(int i=0;i<n;i++) G[i].clear();
        edges.clear();
    }
    void clear_flow(){
        int len=edges.size();
        for(int i=0;i<len;i++) edges[i].flow=0;
    }
    void add_edge(int from,int to,int cap){
        edges.push_back((Edge){from,to,cap,0});
        edges.push_back((Edge){to,from,0,0});
        m=edges.size();
        G[from].push_back(m-2);
        G[to].push_back(m-1);
    }
    bool BFS(){
        memset(vis,0,sizeof(vis));
        queue<int> q;
        q.push(s);
        d[s]=0;
        vis[s]=1;
        while(!q.empty()){
            int x=q.front();
            q.pop();
            int len=G[x].size();
            for(int i=0;i<len;i++){
                Edge& e=edges[G[x][i]];
                if(!vis[e.to]&&e.cap>e.flow){
                    vis[e.to]=1;
                    d[e.to]=d[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,len=G[x].size();
        for(int& i=cur[x];i<len;i++){
            Edge& e=edges[G[x][i]];
            if(d[x]+1==d[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;
    }
    int mincut(){   //call this after maxflow
        int ans=0;
        int len=edges.size();
        for(int i=0;i<len;i++){
            Edge& e=edges[i];
            if(vis[e.from]&&!vis[e.to]&&e.cap>0) ans++;
        }
        return ans;
    }
    void reduce(){
        int len=edges.size();
        for(int i=0;i<len;i++) edges[i].cap-=edges[i].flow;
    }
}solver;

int d[MAXN];
vector<int> p[MAXN];

struct HeapNode{
    int u,d;
    bool operator < (const HeapNode& rhs) const{
        return d>rhs.d;
    }
};
struct Edge2{
    int u,v,dist;
};

struct Dijkstra{
    int n,m;
    vector<Edge2> edges;
    vector<int> G[MAXN];
    bool done[MAXN];

    void init(int n){
        this->n=n;
        for(int i=0;i<n;i++) G[i].clear();
        edges.clear();
    }
    void add_edge(int u,int v,int dist){
        edges.push_back((Edge2){u,v,dist});
        m=edges.size();
        G[u].push_back(m-1);
    }
    void dijkstra(int s){
        priority_queue<HeapNode> q;
        for(int i=0;i<n;i++) p[i].clear();
        for(int i=0;i<n;i++) d[i]=INF;
        d[s]=0;
        memset(done,0,sizeof(done));
        q.push((HeapNode){s,0});
        while(!q.empty()){
            HeapNode x=q.top();
            q.pop();
            int u=x.u;
            if(done[u]) continue;
            done[u]=true;
            int L=G[u].size();
            for(int i=0;i<L;i++){
                Edge2& e=edges[G[u][i]];
                if(d[e.v]>d[u]+e.dist){
                    d[e.v]=d[u]+e.dist;
                    p[e.v].clear();
                    p[e.v].push_back(u);
                    q.push((HeapNode){e.v,d[e.v]});
                }
                else if(d[e.v]==d[u]+e.dist){
                    p[e.v].push_back(u);
                    q.push((HeapNode){e.v,d[e.v]});
                }
            }
        }
    }
}solver2;

int num[MAXN];

int dfs(int u){
    if(u==0) return 0;
    if(num[u]!=INF) return num[u];
    int len=p[u].size();
    for(int i=0;i<len;i++){
        int fa=p[u][i];
        solver.add_edge(fa,u,1);
        solver.add_edge(u,fa,1);
        num[u]=min(num[u],dfs(fa)+1);
    }
    return num[u];
}

int main(){
    freopen("in.txt","r",stdin);
    while(scanf("%d%d",&N,&M)!=EOF){
        solver2.init(N+1);
        int u,v,t;
        for(int i=0;i<M;i++){
            scanf("%d%d%d",&u,&v,&t);
            u--;
            v--;
            solver2.add_edge(u,v,t);
            solver2.add_edge(v,u,t);
        }
        solver2.dijkstra(0);
        memset(num,INF,sizeof(num));
        solver.clear_all(N+1);
        int n=dfs(N-1);
        solver.maxflow(0,N-1);
        int ans1=solver.mincut(),ans2=M-num[N-1];
        printf("%d %d
",ans1,ans2);
    }
    return 0;
}

JAva呼び出し.net webservice

EverBlogを使ってブログ|Hexoにイメージノートを素早く投稿