hdu2255：奔小康赚大钱（二分图带权最大匹配）

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 17911 Accepted Submission(s): 7534

Problem Description

传说在遥远的地方有一个非常富裕的村落,有一天,村长决定进行制度改革：重新分配房子。
这可是一件大事,关系到人民的住房问题啊。村里共有n间房间,刚好有n家老百姓,考虑到每家都要有房住（如果有老百姓没房子住的话，容易引起不安定因素），每家必须分配到一间房子且只能得到一间房子。
另一方面,村长和另外的村领导希望得到最大的效益,这样村里的机构才会有钱.由于老百姓都比较富裕,他们都能对每一间房子在他们的经济范围内出一定的价格,比如有3间房子,一家老百姓可以对第一间出10万,对第2间出2万,对第3间出20万.(当然是在他们的经济范围内).现在这个问题就是村领导怎样分配房子才能使收入最大.(村民即使有钱购买一间房子但不一定能买到,要看村领导分配的).

Input

输入数据包含多组测试用例，每组数据的第一行输入n,表示房子的数量(也是老百姓家的数量)，接下来有n行,每行n个数表示第i个村名对第j间房出的价格(n<=300)。

Output

请对每组数据输出最大的收入值，每组的输出占一行。

Sample Input

 
100 10 15 23 
 

Sample Output

123

Source

HDOJ 2008 Summer Exercise（4）- Buffet Dinner

Recommend

lcy | We have carefully selected several similar problems for you: 2063 1068 3360 1083 2444

#include <iostream>
#include <cstring>
#include <cstdio>

using namespace std;
const int MAXN = 305;
const int INF = 0x3f3f3f3f;

int love[MAXN][MAXN];   // 记录每个妹子和每个男生的好感度
int ex_girl[MAXN];      // 每个妹子的期望值
int ex_boy[MAXN];       // 每个男生的期望值
bool vis_girl[MAXN];    // 记录每一轮匹配匹配过的女生
bool vis_boy[MAXN];     // 记录每一轮匹配匹配过的男生
int match[MAXN];        // 记录每个男生匹配到的妹子 如果没有则为-1
int slack[MAXN];        // 记录每个汉子如果能被妹子倾心最少还需要多少期望值

int N;


bool dfs(int girl)
{
    vis_girl[girl] = true;

    for (int boy = 0; boy < N; ++boy) {

        if (vis_boy[boy]) continue; // 每一轮匹配 每个男生只尝试一次

        int gap = ex_girl[girl] + ex_boy[boy] - love[girl][boy];

        if (gap == 0) {  // 如果符合要求
            vis_boy[boy] = true;
            if (match[boy] == -1 || dfs( match[boy] )) {    // 找到一个没有匹配的男生 或者该男生的妹子可以找到其他人
                match[boy] = girl;
                return true;
            }
        } else {
            slack[boy] = min(slack[boy], gap);  // slack 可以理解为该男生要得到女生的倾心 还需多少期望值 取最小值 备胎的样子【捂脸
        }
    }

    return false;
}

int KM()
{
    memset(match, -1, sizeof match);    // 初始每个男生都没有匹配的女生
    memset(ex_boy, 0, sizeof ex_boy);   // 初始每个男生的期望值为0

    // 每个女生的初始期望值是与她相连的男生最大的好感度
    for (int i = 0; i < N; ++i) {
        ex_girl[i] = love[i][0];
        for (int j = 1; j < N; ++j) {
            ex_girl[i] = max(ex_girl[i], love[i][j]);
        }
    }

    // 尝试为每一个女生解决归宿问题
    for (int i = 0; i < N; ++i) {

        fill(slack, slack + N, INF);    // 因为要取最小值 初始化为无穷大

        while (1) {
            // 为每个女生解决归宿问题的方法是 ：如果找不到就降低期望值，直到找到为止

            // 记录每轮匹配中男生女生是否被尝试匹配过
            memset(vis_girl, false, sizeof vis_girl);
            memset(vis_boy, false, sizeof vis_boy);

            if (dfs(i)) break;  // 找到归宿 退出

            // 如果不能找到 就降低期望值
            // 最小可降低的期望值
            int d = INF;
            for (int j = 0; j < N; ++j)
                if (!vis_boy[j]) d = min(d, slack[j]);

            for (int j = 0; j < N; ++j) {
                // 所有访问过的女生降低期望值
                if (vis_girl[j]) ex_girl[j] -= d;

                // 所有访问过的男生增加期望值
                if (vis_boy[j]) ex_boy[j] += d;
                // 没有访问过的boy 因为girl们的期望值降低，距离得到女生倾心又进了一步！
                else slack[j] -= d;
            }
        }
    }

    // 匹配完成 求出所有配对的好感度的和
    int res = 0;
    for (int i = 0; i < N; ++i)
        res += love[ match[i] ][i];

    return res;
}

int main()
{
    while (~scanf("%d", &N)) {

        for (int i = 0; i < N; ++i)
            for (int j = 0; j < N; ++j)
                scanf("%d", &love[i][j]);

        printf("%d\n", KM());
    }
    return 0;
}

Vasya and Endless Credits

CodeForces - 1107F

Vasya wants to buy himself a nice new car. Unfortunately, he lacks some money. Currently he has exactly 0 burles.

However, the local bank has $n$

credit offers. Each offer can be described with three numbers

a_{i}

months (including the month he activated the offer). Vasya can take the offers any order he wants.

扫描二维码关注公众号，回复： 7478276 查看本文章

Each month Vasya can take no more than one credit offer. Also each credit offer can not be used more than once. Several credits can be active at the same time. It implies that Vasya pays bank the sum of $b_{i}$

a_{i}

of active credits at the end of each month.

Vasya wants to buy a car in the middle of some month. He just takes all the money he currently has and buys the car of that exact price.

Vasya don't really care what he'll have to pay the bank back after he buys a car. He just goes out of the country on his car so that the bank can't find him anymore.

What is the maximum price that car can have?

a_{i}

The first line contains one integer $n$

a_{i}

) — the number of credit offers.

Each of the next $n$

a_{i}

a_{i}

Print one integer — the maximum price of the car.

a_{i}

Input

Output

Input

3
40 1 2
1000 1100 5
300 2 1

Output

a_{i}

In the first example, the following sequence of offers taken is optimal: 4 $\to$

The amount of burles Vasya has changes the following way: 5 $\to$

a_{i}

.... He takes the money he has in the middle of the second month (32 burles) and buys the car.

The negative amount of money means that Vasya has to pay the bank that amount of burles.

In the second example, the following sequence of offers taken is optimal: 3 $\to$

a_{i}

The amount of burles Vasya has changes the following way: 0 $\to$

a_{i}

#include <iostream>
#include <cstring>
#include <cstdio>
typedef long long ll;
using namespace std;
const int MAXN = 605;
const ll INF = 0x3f3f3f3f3f3f3f;

ll love[MAXN][MAXN];   // 记录每个妹子和每个男生的好感度
ll ex_girl[MAXN];      // 每个妹子的期望值
ll ex_boy[MAXN];       // 每个男生的期望值
bool vis_girl[MAXN];    // 记录每一轮匹配匹配过的女生
bool vis_boy[MAXN];     // 记录每一轮匹配匹配过的男生
int match[MAXN];        // 记录每个男生匹配到的妹子 如果没有则为-1
ll slack[MAXN];        // 记录每个汉子如果能被妹子倾心最少还需要多少期望值

int N;

ll max(ll a,ll b){
    return a>b?a:b;
}

ll min(ll a,ll b){
    return a<b?a:b;
}

bool dfs(ll girl)
{
    vis_girl[girl] = true;

    for (int boy = 0; boy < N; ++boy) {

        if (vis_boy[boy]) continue; // 每一轮匹配 每个男生只尝试一次

        ll gap = ex_girl[girl] + ex_boy[boy] - love[girl][boy];

        if (gap == 0) {  // 如果符合要求
            vis_boy[boy] = true;
            if (match[boy] == -1 || dfs( match[boy] )) {    // 找到一个没有匹配的男生 或者该男生的妹子可以找到其他人
                match[boy] = girl;
                return true;
            }
        } else {
            slack[boy] = min(slack[boy], gap);  // slack 可以理解为该男生要得到女生的倾心 还需多少期望值 取最小值 备胎的样子【捂脸
        }
    }

    return false;
}

ll KM()
{
    memset(match, -1, sizeof match);    // 初始每个男生都没有匹配的女生
    memset(ex_boy, 0, sizeof ex_boy);   // 初始每个男生的期望值为0

    // 每个女生的初始期望值是与她相连的男生最大的好感度
    for (int i = 0; i < N; ++i) {
        ex_girl[i] = love[i][0];
        for (int j = 1; j < N; ++j) {
            ex_girl[i] = max(ex_girl[i], love[i][j]);
        }
    }

    // 尝试为每一个女生解决归宿问题
    for (int i = 0; i < N; ++i) {

        fill(slack, slack + N, INF);    // 因为要取最小值 初始化为无穷大

        while (1) {
            // 为每个女生解决归宿问题的方法是 ：如果找不到就降低期望值，直到找到为止

            // 记录每轮匹配中男生女生是否被尝试匹配过
            memset(vis_girl, false, sizeof vis_girl);
            memset(vis_boy, false, sizeof vis_boy);

            if (dfs(i)) break;  // 找到归宿 退出

            // 如果不能找到 就降低期望值
            // 最小可降低的期望值
            ll d = INF;
            for (int j = 0; j < N; ++j)
                if (!vis_boy[j]) d = min(d, slack[j]);

            for (int j = 0; j < N; ++j) {
                // 所有访问过的女生降低期望值
                if (vis_girl[j]) ex_girl[j] -= d;

                // 所有访问过的男生增加期望值
                if (vis_boy[j]) ex_boy[j] += d;
                // 没有访问过的boy 因为girl们的期望值降低，距离得到女生倾心又进了一步！
                else slack[j] -= d;
            }
        }
    }

    // 匹配完成 求出所有配对的好感度的和
    ll res = 0;
    for (int i = 0; i < N; ++i)
        res += love[ match[i] ][i];

    return res;
}

int main()
{
    while (~scanf("%d", &N)) {
        for(int i=0;i<N;i++){
            ll ta,tb,tc;
            scanf("%lld%lld%lld",&ta,&tb,&tc);            
            for(int j=0;j<N;j++){
                ll tem=max(ta-(ll)tb*min(tc,j),0);
                    love[i][j]=tem;
            } 
        }
        cout<<KM()<<endl;
    }
    return 0;
}

二分图相关知识

hdu2255：奔小康赚大钱（二分图带权最大匹配）

Vasya and Endless Credits

猜你喜欢