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  • Luogu[POI2005]KOS-Dicing

    题面


    二分后用网络流判定
    S->人,流量为二分的mid
    人->比赛,流量为1
    比赛->T,流量为1
    输出方案只要判断a就可以了


    # include <bits/stdc++.h>
    # define IL inline
    # define RG register
    # define Fill(a, b) memset(a, b, sizeof(a))
    # define Copy(a, b) memcpy(a, b, sizeof(a))
    # define ID(a, b) n * (a - 1) + b
    using namespace std;
    typedef long long ll;
    const int _(2e4 + 10), __(2e5 + 10), INF(2147483647);
    
    IL ll Read(){
        RG char c = getchar(); RG ll x = 0, z = 1;
        for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
        for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
        return x * z;
    }
    
    int n, m, a[_], w[__], fst[_], nxt[__], to[__], cnt, S, T, lev[_], cur[_], max_flow, tmp1[_], tmp2[__], tcnt, ans[_];
    queue <int> Q;
    
    IL void Add(RG int u, RG int v, RG int f){
        w[cnt] = f; to[cnt] = v; nxt[cnt] = fst[u]; fst[u] = cnt++;
        w[cnt] = 0; to[cnt] = u; nxt[cnt] = fst[v]; fst[v] = cnt++;
    }
    
    IL int Dfs(RG int u, RG int maxf){
        if(u == T) return maxf;
        RG int ret = 0;
        for(RG int &e = cur[u]; e != -1; e = nxt[e]){
            if(lev[to[e]] != lev[u] + 1 || !w[e]) continue;
            RG int f = Dfs(to[e], min(w[e], maxf - ret));
            ret += f; w[e ^ 1] += f; w[e] -= f;
            if(ret == maxf) break;
        }
        if(!ret) lev[u] = 0;
        return ret;
    }
    
    IL bool Bfs(){
        Fill(lev, 0); lev[S] = 1; Q.push(S);
        while(!Q.empty()){
            RG int u = Q.front(); Q.pop();
            for(RG int e = fst[u]; e != -1; e = nxt[e]){
                if(lev[to[e]] || !w[e]) continue;
                lev[to[e]] = lev[u] + 1;
                Q.push(to[e]);
            }
        }
        return lev[T];
    }
    
    IL void Getans(){
        for(RG int i = 1; i <= m; i++)
            for(RG int e = fst[i + n]; e != -1; e = nxt[e])
                if(to[e] && to[e] <= n && w[e]){
                    if(to[e] == a[i]) ans[i] = 1;
                    else ans[i] = 0;
                }
    }
    
    IL bool Check(RG int x){
        Copy(fst, tmp1); Copy(w, tmp2); cnt = tcnt;
        for(RG int i = 1; i <= n; i++) Add(S, i, x);
        for(max_flow = 0; Bfs(); ) Copy(cur, fst), max_flow += Dfs(S, INF);
        return max_flow == m;
    }
    
    int main(RG int argc, RG char* argv[]){
        Fill(fst, -1); n = Read(); m = Read(); T = n + m + 1;
        for(RG int i = 1, b; i <= m; i++){
            a[i] = Read(), b = Read();
            Add(a[i], i + n, 1); Add(b, i + n, 1); Add(i + n, T, 1);
        }
        Copy(tmp1, fst); Copy(tmp2, w); tcnt = cnt;
        RG int l = 1, r = m, num = 0;
        while(l <= r){
            RG int mid = (l + r) >> 1;
            if(Check(mid)) r = mid - 1, num = mid, Getans();
            else l = mid + 1;
        }
        printf("%d
    ", num);
        for(RG int i = 1; i <= m; i++) printf("%d
    ", ans[i]);
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/cjoieryl/p/8206346.html
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