jzoj2701 【GDKOI2012模拟02.01】矩阵

传送门：https://jzoj.net/senior/#main/show/2701

【题目大意】

给出矩阵A，求矩阵B，使得

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" alt="" width="240" height="124" />

最小，矩阵B每个元素在[L,R]内

n<=200,1<=Aij,L,R<=1000, L<=R

【题解】

我们二分答案x，然后对行列建点，每行向每列连[L,R]的边，然后S向每行连[max(S[i]-x), S[i]+x]的边，每列向T连[max(T[i]-x), T[i]+x]的边，判断是否有可行流即可。

其中S[i]表示A中第i行的和，T[i]表示A中第i列的和。

判断有源汇上下界可行流：按照无源汇那样建边，多来一条(T->S，[0,inf])即可。

然后跑dinic，判断从SS流出的边是否全流满。

# include <queue>
# include <stdio.h>
# include <string.h>
# include <iostream>
# include <algorithm>
// # include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
const int M =  + , N = 5e5 + , inf = 1e9;
const int mod = 1e9+;

# define RG register
# define ST static

int n, m, a[M][M], s1[M], s2[M], S, T, L, R;
int head[M * ], nxt[N], to[N], flow[N], tot, indeg[M * ];
inline void add(int u, int v, int fl) {
    ++tot; nxt[tot] = head[u]; head[u] = tot;
    to[tot] = v; flow[tot] = fl;
}
inline void adde(int u, int v, int fl) {
    add(u, v, fl), add(v, u, );
}

inline void tadd(int u, int v, int lf, int rf) {
    adde(u, v, rf-lf);
    indeg[v] += lf;
    indeg[u] -= lf;
}

namespace MF {
    queue<int> q;

    int c[M * ], cur[M * ];
    inline bool bfs() {
        for (int i=; i<=n+m+; ++i) c[i] = -;
        while(!q.empty()) q.pop();
        q.push(S); c[S] = ;
        while(!q.empty()) {
            int top = q.front(); q.pop();
            for (int i=head[top]; i; i=nxt[i]) {
                if(c[to[i]] != - || flow[i] == ) continue;
                c[to[i]] = c[top] + ;
                q.push(to[i]);
                if(to[i] == T) return ;
            }
        }
        return ;
    }

    inline int dfs(int x, int low) {
        if(x == T) return low;
        int r = low, fl;
        for (int i=cur[x]; i; i=nxt[i]) {
            if(c[to[i]] != c[x]+ || flow[i] == ) continue;
            fl = dfs(to[i], min(r, flow[i]));
            flow[i] -= fl; flow[i^] += fl; r -= fl;
            if(flow[i] > ) cur[x] = i;
            if(!r) return low;
        }
        if(r == low) c[x] = -;
        return low-r;
    }

    inline int main() {
        int ans = ;
        while(bfs()) {
            for (int i=; i<=n+m+; ++i) cur[i] = head[i];
            ans += dfs(S, inf);
        }
        return ans;
    }
}

# define line(x) (x)
# define row(x) (x+n) 

inline void build(int x) {
    int SS = n+m+, TT = n+m+;
    S = n+m+, T = n+m+;
    for (int i=; i<=n+m+; ++i) indeg[i] = ;
    for (int i=; i<=n; ++i)
        for (int j=; j<=m; ++j)
            tadd(line(i), row(j), L, R);
    for (int i=; i<=n; ++i)
        tadd(SS, line(i), max(, s1[i]-x), s1[i]+x);
    for (int i=; i<=m; ++i)
        tadd(row(i), TT, max(, s2[i]-x), s2[i]+x);
    tadd(TT, SS, , inf);
    for (int i=; i<=n+m+; ++i) {
        if(indeg[i] < ) adde(i, T, -indeg[i]);
        else adde(S, i, indeg[i]);
    }
}

inline bool chk(int x) {
    tot = ; memset(head, , sizeof head);
    build(x);
    MF::main();
//    cout << x << endl;
    for (int i=head[S]; i; i=nxt[i]) {
//        cout << flow[i] << ' ';
        if(flow[i] != ) {
//            cout << endl;
            return false;
        }
    }
//    cout << endl;
    return true;
}

inline void gans(int ans) {
    for (int i=; i<=n; ++i, puts("")) {
        for (int j=; j<=m; ++j) {
            for (int p=head[line(i)]; p; p=nxt[p]) {
                if(to[p] == row(j)) {
                    printf("%d ", flow[p^] + L);
                    break;
                }
            }
        }
    }
}

int main() {
    freopen("mat.in", "r", stdin);
    freopen("mat.out", "w", stdout);
    cin >> n >> m;
    for (int i=; i<=n; ++i)
        for (int j=; j<=m; ++j) {
            scanf("%d", &a[i][j]);
            s1[i] += a[i][j];
            s2[j] += a[i][j];
        }
    cin >> L >> R;
    int l = , r = 2e5, mid, ans = ;
    while() {
        if(r-l <= ) {
            for (int i=l; i<=r; ++i)
                if(chk(i)) {
                    ans = i;
                    break;
                }
            break;
        }
        mid = l+r>>;
        if(chk(mid)) r = mid;
        else l = mid;
    }
    cout << ans << endl;
    gans(ans);
    return ;
}