http://fastvj.rainng.com/contest/236779#problem/G

Description：

　　n 行 m 列

　　给你行和 与 列和

　　然后有Q个限制，表示特定单元格元素大小的范围，最后问你可行的矩阵值

Solution：
　　有源汇上下界最大流问题，初始源点 连 行和流量是该行对应得行和，然后列连初始汇点，容量为列的列和，详细的限制，对应于行列之间，因为我最后要输出一个矩阵，所以n * m每一条边都要链接最后，手动连一条 t s inf的边，变成无源汇有上下界可行流问题，然后拆边，把对应边的流量，放到数组里，输出就好啦

Code：

　　前面比较基础的Dinic，数据操作函数，加边操作

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

#include <queue>

#define inf (1 << 28)

using namespace std;

const int maxn = 25;

const int maxm = 210;

const int mn = 505;

const int mm = 440020;

struct node{

    int to,val,pre,lid;

}e[mm];

int id[mn],cnt=0;

int cur[mn];

int flor[mn];

int upflow[mn];

int lowf[maxm][maxn];

int upf[maxm][maxn];

int out[maxm][maxn];

void init()

{

    memset(id,-1,sizeof(id));

    memset(upflow,0,sizeof(upflow));

    cnt = 0;

}

void add(int from,int to,int val,int lid)

{

    e[cnt].to = to;

    e[cnt].val = val;

    e[cnt].lid = lid;

    e[cnt].pre = id[from];

    id[from] = cnt++;

    swap(from,to);

    e[cnt].to = to;

    e[cnt].val = 0;

    e[cnt].lid = lid;

    e[cnt].pre = id[from];

    id[from] = cnt++;

}

void addflow(int from,int to,int low,int up,int lid)

{

    upflow[from] -= low;

    upflow[to] += low;

    add(from,to,up-low,lid);

}

bool bfs(int s,int t)

{

    memset(flor,0,sizeof(flor));

    flor[s] = 1;

    queue<int> q;

    while(q.size())q.pop();

    q.push(s);

    while(q.size())

    {

        int now = q.front();

        q.pop();

        for(int i = id[now];~i;i = e[i].pre)

        {

            int to = e[i].to;

            int val = e[i].val;

            if(val > 0 && flor[to] == 0)

            {

                flor[to] = flor[now] + 1;

                q.push(to);

                if(to == t)

                    return true;

            }

        }

    }

    return false;

}

int dfs(int s,int t,int value)

{

    if(s == t || value == 0)return value;

    int ret = value,a;

    for(int &i = cur[s];~i;i = e[i].pre)

    {

        int val = e[i].val;

        int to = e[i].to;

        if(flor[to] == flor[s] + 1 && (a = dfs(to,t,min(ret,val))))

        {

            e[i].val -= a;

            e[i^1].val += a;

            ret -= a;

            if(ret == 0)break;

        }

    }

    if(ret == value)flor[s] = 0;

    return value - ret;

}

int Dinic(int s,int t)

{

    int ret = 0;

    while(bfs(s,t))

    {

        memcpy(cur,id,sizeof(id));

        ret += dfs(s,t,inf);

    }

    return ret;

}

void addlimit(int i,int j,char op,int lim)

{

    if(op == '=')

    {

        upf[i][j] = lowf[i][j] = lim;

    }

    else if(op == '>')

    {

        lowf[i][j] = max(lowf[i][j],lim+1);

    }

    else

    {

        upf[i][j] = min(upf[i][j],lim-1);

    }

}

然后根据输入一点点的加边，这是行和和列和对应的边

scanf("%d%d",&n,&m);

        s = 0;

        t = n + m + 1;

        ss = t + 1;

        tt = ss + 1;

        int lsum;

        for(int i = 1;i <= n;++i)

        {

            scanf("%d",&lsum);

            addflow(s,i,lsum,lsum,0);

        }

        for(int i = 1;i <= m;++i)

        {

            scanf("%d",&lsum);

            addflow(n + i,t,lsum,lsum,0);

        }

        int limitnum;

然后根据题目中给出的限制条件，填充上下界数组

scanf("%d",&limitnum);

        for(int i = 1;i <= n;++i)

        {

            for(int j = 1;j <= m;++j)

            {

                lowf[i][j] = 0;

                upf[i][j] = inf;

            }

        }

        int row,col,lim;

        char op;

        for(int i = 1;i <= limitnum;++i)

        {

            scanf("%d %d %c %d",&row,&col,&op,&lim);

            if(row == 0 && col == 0)

            {

                for(int i = 1;i <= n;++i)

                {

                    for(int j = 1;j <= m;++j)

                    {

                        addlimit(i,j,op,lim);

                    }

                }

            }

            else if(row == 0)

            {

                for(int i = 1;i <= n;++i)

                {

                    addlimit(i,col,op,lim);

                }

            }

            else if(col == 0)

            {

                for(int i = 1;i <= m;++i)

                {

                    addlimit(row,i,op,lim);

                }

            }

            else

            {

                addlimit(row,col,op,lim);

            }

        }

根据上下界数组，添加有上下界边

int tot = 0;

        for(int i = 1;i <= n;++i)

        {

            for(int j = 1;j <= m;++j)

            {

                addflow(i,n+j,lowf[i][j],upf[i][j],++tot);

            }

        }

加边完成后，转化为无源汇有上下界可行流 ———— 循环流

add(t,s,inf,0);

        int sum = 0;

        for(int i = s;i <= t;++i)

        {

            if(upflow[i] < 0)

            {

                add(i,tt,-upflow[i],0);

            }

            else

            {

                sum += upflow[i];

                add(ss,i,upflow[i],0);

            }

        }

跑DIinc，如果存在可行流的话

就输出叭~~

for(int now = n+1;now <= n + m;++now)

            {

                for(int i = id[now];~i;i = e[i].pre)

                {

                    int to = e[i].to;

                    int lid = e[i].lid;

                    if(lid == 0 || i % 2 == 0)continue;

                    out[to][now-n] = lowf[to][now-n] + e[i].val;

                }

            }

            for(int i = 1;i <= n;++i)

            {

                for(int j = 1;j <= m;++j)

                {

                    if(j == 1)

                        printf("%d",out[i][j]);

                    else

                        printf(" %d",out[i][j]);

                }

                printf("\n");

            }

完整代码

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

#include <queue>

#define inf (1 << 28)

using namespace std;

const int maxn = 25;

const int maxm = 210;

const int mn = 505;

const int mm = 440020;

struct node{

    int to,val,pre,lid;

}e[mm];

int id[mn],cnt=0;

int cur[mn];

int flor[mn];

int upflow[mn];

int lowf[maxm][maxn];

int upf[maxm][maxn];

int out[maxm][maxn];

void init()

{

    memset(id,-1,sizeof(id));

    memset(upflow,0,sizeof(upflow));

    cnt = 0;

}

void add(int from,int to,int val,int lid)

{

    e[cnt].to = to;

    e[cnt].val = val;

    e[cnt].lid = lid;

    e[cnt].pre = id[from];

    id[from] = cnt++;

    swap(from,to);

    e[cnt].to = to;

    e[cnt].val = 0;

    e[cnt].lid = lid;

    e[cnt].pre = id[from];

    id[from] = cnt++;

}

void addflow(int from,int to,int low,int up,int lid)

{

    upflow[from] -= low;

    upflow[to] += low;

    add(from,to,up-low,lid);

}

bool bfs(int s,int t)

{

    memset(flor,0,sizeof(flor));

    flor[s] = 1;

    queue<int> q;

    while(q.size())q.pop();

    q.push(s);

    while(q.size())

    {

        int now = q.front();

        q.pop();

        for(int i = id[now];~i;i = e[i].pre)

        {

            int to = e[i].to;

            int val = e[i].val;

            if(val > 0 && flor[to] == 0)

            {

                flor[to] = flor[now] + 1;

                q.push(to);

                if(to == t)

                    return true;

            }

        }

    }

    return false;

}

int dfs(int s,int t,int value)

{

    if(s == t || value == 0)return value;

    int ret = value,a;

    for(int &i = cur[s];~i;i = e[i].pre)

    {

        int val = e[i].val;

        int to = e[i].to;

        if(flor[to] == flor[s] + 1 && (a = dfs(to,t,min(ret,val))))

        {

            e[i].val -= a;

            e[i^1].val += a;

            ret -= a;

            if(ret == 0)break;

        }

    }

    if(ret == value)flor[s] = 0;

    return value - ret;

}

int Dinic(int s,int t)

{

    int ret = 0;

    while(bfs(s,t))

    {

        memcpy(cur,id,sizeof(id));

        ret += dfs(s,t,inf);

    }

    return ret;

}

void addlimit(int i,int j,char op,int lim)

{

    if(op == '=')

    {

        upf[i][j] = lowf[i][j] = lim;

    }

    else if(op == '>')

    {

        lowf[i][j] = max(lowf[i][j],lim+1);

    }

    else

    {

        upf[i][j] = min(upf[i][j],lim-1);

    }

}

int main()

{

    int T;

    scanf("%d",&T);

    int n,m,s,t,ss,tt;

    while(T--)

    {

        init();

        scanf("%d%d",&n,&m);

        s = 0;

        t = n + m + 1;

        ss = t + 1;

        tt = ss + 1;

        int lsum;

        for(int i = 1;i <= n;++i)

        {

            scanf("%d",&lsum);

            addflow(s,i,lsum,lsum,0);

        }

        for(int i = 1;i <= m;++i)

        {

            scanf("%d",&lsum);

            addflow(n + i,t,lsum,lsum,0);

        }

        int limitnum;

        scanf("%d",&limitnum);

        for(int i = 1;i <= n;++i)

        {

            for(int j = 1;j <= m;++j)

            {

                lowf[i][j] = 0;

                upf[i][j] = inf;

            }

        }

        int row,col,lim;

        char op;

        for(int i = 1;i <= limitnum;++i)

        {

            scanf("%d %d %c %d",&row,&col,&op,&lim);

            if(row == 0 && col == 0)

            {

                for(int i = 1;i <= n;++i)

                {

                    for(int j = 1;j <= m;++j)

                    {

                        addlimit(i,j,op,lim);

                    }

                }

            }

            else if(row == 0)

            {

                for(int i = 1;i <= n;++i)

                {

                    addlimit(i,col,op,lim);

                }

            }

            else if(col == 0)

            {

                for(int i = 1;i <= m;++i)

                {

                    addlimit(row,i,op,lim);

                }

            }

            else

            {

                addlimit(row,col,op,lim);

            }

        }

        int tot = 0;

        for(int i = 1;i <= n;++i)

        {

            for(int j = 1;j <= m;++j)

            {

                addflow(i,n+j,lowf[i][j],upf[i][j],++tot);

            }

        }

        add(t,s,inf,0);

        int sum = 0;

        for(int i = s;i <= t;++i)

        {

            if(upflow[i] < 0)

            {

                add(i,tt,-upflow[i],0);

            }

            else

            {

                sum += upflow[i];

                add(ss,i,upflow[i],0);

            }

        }

        if(Dinic(ss,tt) == sum)

        {

            for(int now = n+1;now <= n + m;++now)

            {

                for(int i = id[now];~i;i = e[i].pre)

                {

                    int to = e[i].to;

                    int lid = e[i].lid;

                    if(lid == 0 || i % 2 == 0)continue;

                    out[to][now-n] = lowf[to][now-n] + e[i].val;

                }

            }

            for(int i = 1;i <= n;++i)

            {

                for(int j = 1;j <= m;++j)

                {

                    if(j == 1)

                        printf("%d",out[i][j]);

                    else

                        printf(" %d",out[i][j]);

                }

                printf("\n");

            }

        }

        else

        {

            printf("IMPOSSIBLE\n");

        }

        if(T)printf("\n");

    }

    return 0;

}

ZOJ1994有源汇上下界可行流的更多相关文章

  1. POJ2396 Budget [有源汇上下界可行流]

    POJ2396 Budget 题意:n*m的非负整数矩阵,给出每行每列的和,以及一些约束关系x,y,>=<,val,表示格子(x,y)的值与val的关系,0代表整行/列都有这个关系,求判断 ...

  2. 有源汇上下界可行流(POJ2396)

    题意:给出一个n*m的矩阵的每行和及每列和,还有一些格子的限制,求一组合法方案. 源点向行,汇点向列,连一条上下界均为和的边. 对于某格的限制,从它所在行向所在列连其上下界的边. 求有源汇上下界可行流 ...

  3. 计蒜客 31447 - Fantastic Graph - [有源汇上下界可行流][2018ICPC沈阳网络预赛F题]

    题目链接:https://nanti.jisuanke.com/t/31447 "Oh, There is a bipartite graph.""Make it Fan ...

  4. poj2396 Budget（有源汇上下界可行流）

    [题目链接] http://poj.org/problem?id=2396 [题意] 知道一个矩阵的行列和,且知道一些格子的限制条件,问一个可行的方案. [思路] 设行为X点,列为Y点,构图:连边(s ...

  5. 算法复习——有源汇上下界可行流（bzoj2396）

    题目: Description We are supposed to make a budget proposal for this multi-site competition. The budge ...

  6. poj2396有源汇上下界可行流

    题意:给一些约束条件,要求算能否有可行流,ps:刚开始输入的是每一列和,那么就建一条上下界相同的边,这样满流的时候就一定能保证流量相同了,还有0是该列(行)对另一行每个点都要满足约束条件 解法:先按无 ...

  7. bzoj 2406 矩阵 —— 有源汇上下界可行流

    题目:https://www.lydsy.com/JudgeOnline/problem.php?id=2406 这题,首先把题目那个式子的绝对值拆成两个限制,就成了网络流的上下界: 有上下界可行流原 ...

  8. bzoj千题计划158：bzoj2406: 矩阵（有源汇上下界可行流）

    http://www.lydsy.com/JudgeOnline/problem.php?id=2406 设矩阵C=A-B 最小化 C 一行或一列和的最大值 整体考虑一行或者一列的和 二分最大值 这样 ...

  9. poj2396 Budget&&ZOJ1994 Budget[有源汇上下界可行流]

    Budget Time Limit: 5 Seconds      Memory Limit: 32768 KB      Special Judge We are supposed to make ...

随机推荐

  1. 获取当前div中的文本(只获取当前div的, 子元素不要, 基于layui)

    <!DOCTYPE html> <html> <head> <meta charset="utf-8"> <title> ...

  2. as3.0 比较两个数组

    var arr1:Array=[1,2,3,4] var arr2:Array=[1,2,4,3] trace(arr1.join(",") == arr2.join(" ...

  3. vim 自动添加作者、版权、修改时间等信息

    相信大家阅读代码时都见过这样的文件头: # THIS FILE IS PART OF LibreBoot PROJECT (归属) # reboot.py - The core part of the ...

  4. PCIE training

    在PCIe链路可以正常工作之前,需要对PCIe链路进行链路训练,在这个过程中,就会用LTSSM状态机.LTSSM全称是Link Training and Status State Machine.这个 ...

  5. MoneyRunner API汇总

    MonkeyRunner API 汇总 MonkeyRunner工具主要有三个类: MonkeyRunner MonkeyDevice MonkeyImage 1.MonkeyRunner类: Mon ...

  6. 53-java中的queue

    java.util 接口 Queue<E> 类型参数: E - collection 中所保存元素的类型. 所有超级接口: Collection<E>, Iterable< ...

  7. Linux驱动之USB鼠标驱动编写

    本篇博客分以下几部分讲解 1.介绍USB四大描述 2.介绍USB鼠标驱动程序功能及框架 3.介绍程序用到的结构体 4.介绍程序用到的函数 5.编写程序 6.测试程序 1.介绍USB四大描述符 USB设 ...

  8. 在MyEclipse中使用spring-boot+mybatis+freemarker实现基本的增删改查

    一.基本环境 二.创建实体类 1.User.java package bjredcross.rainbowplans.model; import bjredcross.rainbowplans.com ...

  9. vue自定义滚动条

    参照element-ui的el-scroll自己实现了一个自定义组件,代码如下: <template> <div class="c-scroll-box" ref ...

  10. leetcode5：最长回文子串

    给定一个字符串 s,找到 s 中最长的回文子串.你可以假设 s 的最大长度为1000. 示例 1: 输入: "babad" 输出: "bab" 注意: &quo ...