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题目大意：给你一个n，m，k。n行，m列。然后给你每一行的总和，与每一列的总和，让你在这个n*m的矩阵里面填一个小于等于k的数字，使得满足每一列，每一行的和。如果没有输出“Impossible”，有多解输出“Not Unique”，有唯一的解输出“Unique”，并输出他的解。

从源点到每一行的和建边容量为它的总和，从汇点到列建边容量为它的总和。然后行到列建边容量为数据上限K。然后求是否存在满流，如果从在在判断是否唯一。判唯一的时候，你要看中间是否存在环流，这样的话有些值 你取多少是不会收到影响的。

Redraw Beautiful Drawings

2 2 4
4 2
4 2  
4 2 2
2 2 5 0
5 4
1 4 3
9
1 2 3 3

Not Unique
Impossible
Unique
1 2 3 3

#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define eps 1e-12
///#define M 1000100
#define LL __int64
///#define LL long long
///#define INF 0x7ffffff
#define INF 0x3f3f3f3f
#define PI 3.1415926535898
#define zero(x) ((fabs(x)<eps)?0:x)

using namespace std;

const int maxn = 1010;

int cnt;
int n, m;
int cur[maxn], head[maxn];
int dis[maxn], gap[maxn];
int aug[maxn], pre[maxn];
int deep[maxn];

struct node
{
    int v, w;
    int next;
} f[500100];
void init()
{
    cnt = 0;
    memset(head, -1, sizeof(head));
}

void add(int u, int v, int w)
{
    f[cnt].v = v;
    f[cnt].w = w;
    f[cnt].next = head[u];
    head[u] = cnt++;

    f[cnt].v = u;
    f[cnt].w = 0;
    f[cnt].next = head[v];
    head[v] = cnt++;
}

int SAP(int s, int e, int n)
{
    int max_flow = 0, v, u = s;
    int id, mindis;
    aug[s] = INF;
    pre[s] = -1;
    memset(dis, 0, sizeof(dis));
    memset(gap, 0, sizeof(gap));
    gap[0] = n;
    for (int i = 0; i <= n; ++i)  cur[i] = head[i];/// 初始化当前弧为第一条弧
    while (dis[s] < n)
    {
        bool flag = false;
        if (u == e)
        {
            max_flow += aug[e];
            for (v = pre[e]; v != -1; v = pre[v]) /// 路径回溯更新残留网络
            {
                id = cur[v];
                f[id].w -= aug[e];
                f[id^1].w += aug[e];
                aug[v] -= aug[e]; /// 修改可增广量，以后会用到
                if (f[id].w == 0) u = v; /// 不回退到源点，仅回退到容量为0的弧的弧尾
            }
        }
        for (id = cur[u]; id != -1; id = f[id].next)/// 从当前弧开始查找允许弧
        {
            v = f[id].v;
            if (f[id].w > 0 && dis[u] == dis[v] + 1) /// 找到允许弧
            {
                flag = true;
                pre[v] = u;
                cur[u] = id;
                aug[v] = min(aug[u], f[id].w);
                u = v;
                break;
            }
        }
        if (flag == false)
        {
            if (--gap[dis[u]] == 0) break; ///gap优化，层次树出现断层则结束算法
            mindis = n;
            cur[u] = head[u];
            for (id = head[u]; id != -1; id = f[id].next)
            {
                v = f[id].v;
                if (f[id].w > 0 && dis[v] < mindis)
                {
                    mindis = dis[v];
                    cur[u] = id; /// 修改标号的同时修改当前弧
                }
            }
            dis[u] = mindis + 1;
            gap[dis[u]]++;
            if (u != s) u = pre[u]; /// 回溯继续寻找允许弧
        }
    }
    return max_flow;
}

int mp[410][410];
int vis[maxn];

bool dfs(int x, int fa)
{
    for(int i = head[x]; i != -1; i = f[i].next)
    {
        if(i == (fa^1)) continue;
        if(!f[i].w) continue;
        if(vis[f[i].v]) return true;
        vis[f[i].v] = 1;
        if(dfs(f[i].v, i)) return true;
        vis[f[i].v] = 0;
    }
    return false;
}

int main()
{
    int k;
    int x, y;
    int S;
    int T;
    int en;
    int sum1;
    int sum2;
    while(~scanf("%d %d %d",&n, &m, &k))
    {
        init();
        sum1 = sum2 = 0;
        S = 0;
        T = n+m+1;
        en = T+1;
        for(int i = 1; i <= n; i++)
        {
            scanf("%d",&x);
            sum1 += x;
            add(S, i, x);
            for(int j = 1; j <= m; j++) add(i, j+n, k);
        }
        for(int i = 1; i <= m; i++)
        {
            scanf("%d",&y);
            sum2 += y;
            add(i+n, T, y);
        }
        if(sum1 != sum2)
        {
            puts("Impossible");
            continue;
        }
        int ans = SAP(S, T, en);
        if(ans != sum1)
        {
            puts("Impossible");
            continue;
        }
        memset(vis, 0, sizeof(vis));
        int flag = 0;
        for(int i = 1; i <= n; i++)
        {
            if(dfs(i, -1))
            {
                flag = 1;
                break;
            }
        }
        if(flag)
        {
            puts("Not Unique");
            continue;
        }
        memset(mp, 0, sizeof(mp));
        puts("Unique");
        for(int i = 1; i <= n; i++)
        {
            for(int j = head[i]; j != -1; j = f[j].next)
            {
                int x = f[j].v;
                if(x > n && x <= n+m) mp[i][x-n] = k-f[j].w;
            }
        }
        for(int i = 1; i <= n; i++)
        {
            for(int j = 1; j < m; j++) printf("%d ",mp[i][j]);
            printf("%d\n",mp[i][m]);
        }
    }
    return 0;
}

HDU 4888 Redraw Beautiful Drawings(最大流),布布扣,bubuko.com

HDU 4888 Redraw Beautiful Drawings(最大流)

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原文地址：http://blog.csdn.net/xu12110501127/article/details/38662287