• 【hihocoder 1369】网络流一·Ford-Fulkerson算法


    Link:http://hihocoder.com/problemset/problem/1369

    Description

    Solution

    最大流模板题

    NumberOf WA


    Reviw


    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 = 500;
    const int INF = 0x3f3f3f3f;
    
    int n,m,flow[N+10][N+10],pre[N+10];
    queue <int> dl;
    
    int main(){
        //Open();
        //Close();
        ri(n),ri(m);
        rep1(i,1,m){
            int x,y,z;
            ri(x),ri(y),ri(z);
            flow[x][y]+=z;
        }
    
        int ans = 0;
        while (1){
            while (!dl.empty()) dl.pop();
            ms(pre,255);
            dl.push(1);
            pre[1] = 0;
            while (!dl.empty()){
                int x = dl.front();
                dl.pop();
                for (int i = 1;i <= n;i++)
                    if (pre[i]==-1 && flow[x][i]){
                        pre[i] = x;
                        dl.push(i);
                    }
            }
            if (pre[n]==-1) break;
            int mi = INF;
            int now = n;
            while (now != 1){
                mi = min(mi,flow[pre[now]][now]);
                now = pre[now];
            }
            now = n;
            while (now != 1){
                flow[pre[now]][now] -= mi;
                now = pre[now];
            }
            ans += mi;
        }
        oi(ans);puts("");
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/AWCXV/p/7626129.html
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