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  • 今日SGU 5.25

    SGU 194

    题意:无源汇有上下界的最大流

    收获:https://wenku.baidu.com/view/0f3b691c59eef8c75fbfb35c.html

    #include<bits/stdc++.h>
    #define de(x) cout<<#x<<"="<<x<<endl;
    #define dd(x) cout<<#x<<"="<<x<<" ";
    #define rep(i,a,b) for(int i=a;i<(b);++i)
    #define repd(i,a,b) for(int i=a;i>=(b);--i)
    #define repp(i,a,b,t) for(int i=a;i<(b);i+=t)
    #define ll long long
    #define mt(a,b) memset(a,b,sizeof(a))
    #define fi first
    #define se second
    #define inf 0x3f3f3f3f
    #define INF 0x3f3f3f3f3f3f3f3f
    #define pii pair<int,int>
    #define pdd pair<double,double>
    #define pdi pair<double,int>
    #define mp(u,v) make_pair(u,v)
    #define sz(a) (int)a.size()
    #define ull unsigned long long
    #define ll long long
    #define pb push_back
    #define PI acos(-1.0)
    #define qc std::ios::sync_with_stdio(false)
    #define db double
    #define all(a) a.begin(),a.end()
    const int mod = 1e9+7;
    const int maxn = 2e2+5;
    const double eps = 1e-6;
    using namespace std;
    bool eq(const db &a, const db &b) { return fabs(a - b) < eps; }
    bool ls(const db &a, const db &b) { return a + eps < b; }
    bool le(const db &a, const db &b) { return eq(a, b) || ls(a, b); }
    ll gcd(ll a,ll b) { return a==0?b:gcd(b%a,a); };
    ll lcm(ll a,ll b) { return a/gcd(a,b)*b; }
    ll kpow(ll a,ll b) {ll res=1;a%=mod; if(b<0) return 1; for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
    ll read(){
        ll x=0,f=1;char ch=getchar();
        while (ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
        while (ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
        return x*f;
    }
    int n,m,s,t,tot;
    int b[maxn*maxn],c[maxn*maxn],in[maxn],out[maxn];
    int pre[maxn],dis[maxn];
    int q[maxn],l,r;
    struct edge{
        int v,w,nt;
    }e[maxn*maxn];
    void init(){
        tot = 0;
        mt(pre,-1);mt(in,0);mt(out,0); 
    }
    void addedge(int u,int v,int w){
        e[tot].v=v;e[tot].w=w;e[tot].nt=pre[u];
        pre[u]=tot++;
        e[tot].v=u;e[tot].w=0;e[tot].nt=pre[v];
        pre[v]=tot++;
    }
    bool bfs(){
        mt(dis,-1);
        dis[s]=0;
        l = r = 0;
        q[r++] = s;
        while(l<r){
            int u=q[l++];
            for(int i=pre[u];~i;i=e[i].nt){
                int v=e[i].v,w=e[i].w;
                if(dis[v]==-1&&w>0){
                    dis[v]=dis[u]+1;
                    q[r++] = v;
                }
            }
        }
        return dis[t]!=-1;
    }
    int dfs(int u,int cp){
        if(u==t||!cp) return cp;
        int tmp=cp;
        for(int i=pre[u];~i;i=e[i].nt){
            int v=e[i].v,w=e[i].w;
            if(dis[v]==dis[u]+1&&w>0){
                int tt=dfs(v,min(tmp,w));
                e[i].w-=tt;
                e[i^1].w+=tt;
                tmp-=tt;
                if(tmp<=0) return cp;
            }
        }
        dis[u]=inf;
        return cp-tmp;
    }
    int dinic(){
        int ret = 0;
        while(bfs()) ret+=dfs(s,inf);
        return ret;
    }
    int main(){
        int u,v;
        init();
        scanf("%d%d",&n,&m);
        rep(i,0,m) scanf("%d%d%d%d",&u,&v,&b[i],&c[i]),addedge(u,v,c[i]-b[i]),in[v] += b[i],out[u] += b[i];
        s = 0;t = n + 1;
        int ans = 0;
        rep(i,1,n+1) if(out[i] - in[i] > 0) addedge(i,t,out[i] - in[i]),ans += out[i] - in[i];
        else addedge(s,i,in[i] - out[i]);
        if(ans!=dinic())  return puts("NO"),0;
        puts("YES");
        rep(i,0,m) printf("%d
    ",b[i] + e[i << 1 | 1].w);
        return 0;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/chinacwj/p/9086779.html
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