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  • 【hdu 3987】Harry Potter and the Forbidden Forest

    Link:http://acm.hdu.edu.cn/showproblem.php?pid=3987

    Description

    给出一张有n个点的图,有的边又向,有的边无向,现在要你破坏一些路,使得从点0无法到达点n-1。破坏每条路都有一个代价。求在代价最小的前提下,最少需要破坏多少条道路。(就是说求在最小割的前提下,最小的割边数)

    Solution

    我们先在原图上跑一次最大流;
    求出跑完最大流之后的剩余网络.
    显然,最后剩余网络上容量变成0的(也就是满流的边);
    它才可能是最小割的边.
    接下来;
    把那些可能是最小割的边的边的容量重新赋值为1;
    然后,其余边都赋值成INF;
    再跑一遍最大流;
    求出此时的最小割.就是答案了.
    也即最小割边数目.

    NumberOf WA

    0

    Reviw

    最小割模型

    Code

    #include <bits/stdc++.h>
    using namespace std;
    #define lson l,m,rt<<1
    #define rson m+1,r,rt<<1|1
    #define LL long long
    #define rep1(i,a,b) for (int i = a;i <= b;i++)
    #define rep2(i,a,b) for (int i = a;i >= b;i--)
    #define mp make_pair
    #define pb push_back
    #define fi first
    #define se second
    #define ms(x,y) memset(x,y,sizeof x)
    #define ri(x) scanf("%d",&x)
    #define rl(x) scanf("%lld",&x)
    #define rs(x) scanf("%s",x+1)
    #define oi(x) printf("%d",x)
    #define ol(x) printf("%lld",x)
    #define oc putchar(' ')
    #define os(x) printf(x)
    #define all(x) x.begin(),x.end()
    #define Open() freopen("F:\rush.txt","r",stdin)
    #define Close() ios::sync_with_stdio(0)
    
    typedef pair<int,int> pii;
    typedef pair<LL,LL> pll;
    
    const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
    const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
    const double pi = acos(-1.0);
    const int N = 1000;
    const int M = 4e5;
    const LL INF = 1e18;
    
    int n,m,en[M+10],fir[N+10],tfir[N+10],nex[M+10],totm;
    int deep[N+10];
    LL flow[M+10];
    queue <int> dl;
    
    void add(int x,int y,int z){
        nex[totm] = fir[x];
        fir[x] = totm;
        flow[totm] = z;
        en[totm] = y;
        totm++;
    
        nex[totm] = fir[y];
        fir[y] = totm;
        flow[totm] = 0;
        en[totm] = x;
        totm++;
    }
    
    bool bfs(){
        ms(deep,255);
        dl.push(1);
        deep[1] = 0;
        while (!dl.empty()){
            int x = dl.front();
            dl.pop();
            for (int i = fir[x];i != -1;i = nex[i]){
                int y = en[i];
                if (flow[i] >0 && deep[y]==-1){
                    deep[y] = deep[x] + 1;
                    dl.push(y);
                }
            }
        }
        return deep[n]!=-1;
    }
    
    LL dfs(int x,LL limit){
        if (x == n) return limit;
        if (limit == 0) return 0;
        LL cur,f = 0;
        for (int i = tfir[x];i!=-1;i = nex[i]){
            tfir[x] = i;
            int y = en[i];
            if (deep[y] == deep[x] + 1 && (cur = dfs(y,min(limit,flow[i])))) {
                limit -= cur;
                f += cur;
                flow[i] -= cur;
                flow[i^1] += cur;
                if (!limit) break;
            }
        }
        return f;
    }
    
    int main(){
        //Open();
        //Close();
        int T,kk = 0;
        ri(T);
        while (T--){
            ms(fir,255);
            totm = 0;
            ri(n),ri(m);
            rep1(i,1,m){
                int x,y,z,d;
                ri(x),ri(y),ri(z),ri(d);
                x++,y++;
                add(x,y,z);
                if (d == 1) add(y,x,z);
            }
            while ( bfs() ) {
                rep1(i,1,n) tfir[i] = fir[i];
                dfs(1,INF);
            }
    
            for (int i = 0;i < totm;i+=2)
                if (flow[i]==0){
                    flow[i] = 1;
                    flow[i^1] = 0;
                }else{
                    flow[i] = INF;
                    flow[i^1] = 0;
                }
    
            LL ans = 0;
    
            while ( bfs() ) {
                rep1(i,1,n) tfir[i] = fir[i];
                ans+=dfs(1,INF);
            }
            os("Case ");oi(++kk);os(": ");ol(ans);puts("");
        }
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/AWCXV/p/7626125.html
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