HDU 6346 整数规划 (最佳完美匹配)

时间：2023-03-08 16:07:56

整数规划

Time Limit: 5500/5000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 435    Accepted Submission(s): 144

Problem Description
度度熊有一个可能是整数规划的问题：

给定 n×n 个整数 ai,j(1≤i,j≤n)，要找出 2n 个整数 x1,x2,…,xn,y1,y2,…,yn 在满足 xi+yj≤ai,j(1≤i,j≤n) 的约束下最大化目标函数 ∑ni=1xi+∑ni=1yi，

你需要帮他解决这个整数规划问题，并给出目标函数的最大值。

Input
第一行包含一个整数 T，表示有 T 组测试数据。

接下来依次描述 T 组测试数据。对于每组测试数据：

第一行包含一个整数 n，表示该整数规划问题的规模。

接下来 n 行，每行包含 n 个整数，其中第 i 行第 j 列的元素是 ai,j。

保证 1≤T≤20，1≤n≤200，−109≤ai,j≤109(1≤i,j≤n)。

Output
对于每组测试数据，输出一行信息 "Case #x: y"（不含引号），其中 x 表示这是第 x 组测试数据，y 表示目标函数的最大值，行末不要有多余空格。
Sample Input
2
1
0
2
1 2
3 4
Sample Output
Case #1: 0
Case #2: 5
Source
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析：根据题目给定的条件 xi+yj≤ai,j(1≤i,j≤n)，这就是求最佳完全匹配（最小权值）的可行顶标的定义，所以直接就是一个裸的 KM 算法，因为是最小权值，所以只要把权值取反即可。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <cstdio>

#include <string>

#include <cstdlib>

#include <cmath>

#include <iostream>

#include <cstring>

#include <set>

#include <queue>

#include <algorithm>

#include <vector>

#include <map>

#include <cctype>

#include <cmath>

#include <stack>

#include <sstream>

#include <list>

#include <assert.h>

#include <bitset>

#include <numeric>

#define debug() puts("++++")

#define gcd(a, b) __gcd(a, b)

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define fi first

#define se second

#define pb push_back

#define sqr(x) ((x)*(x))

#define ms(a,b) memset(a, b, sizeof a)

#define sz size()

#define be begin()

#define ed end()

#define pu push_up

#define pd push_down

#define cl clear()

#define lowbit(x) -x&x

//#define all 1,n,1

#define FOR(i,n,x)  for(int i = (x); i < (n); ++i)

#define freopenr freopen("in.in", "r", stdin)

#define freopenw freopen("out.out", "w", stdout)

using namespace std;

typedef long long LL;

typedef unsigned long long ULL;

typedef pair<int, int> P;

const int INF = 0x3f3f3f3f;

const LL LNF = 1e17;

const double inf = 1e20;

const double PI = acos(-1.0);

const double eps = 1e-8;

const int maxn = 200 + 10;

const int maxm = 1e6 + 10;

const LL mod = 998244353LL;

const int dr[] = {-1, 1, 0, 0, 1, 1, -1, -1};

const int dc[] = {0, 0, 1, -1, 1, -1, 1, -1};

const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};

int n, m;

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

inline bool is_in(int r, int c) {

  return r >= 0 && r < n && c >= 0 && c < m;

}

inline int readInt(){ int x;  scanf("%d", &x);  return x; }

LL w[maxn][maxn], x[maxn], y[maxn], slack[maxn];

int prev_x[maxn], prev_y[maxn], son_y[maxn], par[maxn];

int lx, ly;

void adjust(int v){

  son_y[v] = prev_y[v];

  if(prev_x[son_y[v]] != -2)  adjust(prev_x[son_y[v]]);

}

bool find(int v){

  for(int i = 0; i < n; ++i)  if(prev_y[i] == -1){

    if(slack[i] > x[v] + y[i] - w[v][i]){

      slack[i] = x[v] + y[i] - w[v][i];

      par[i] = v;

    }

    if(x[v] + y[i] == w[v][i]){

      prev_y[i] = v;

      if(son_y[i] == -1){

        adjust(i);  return true;

      }

      if(prev_x[son_y[i]] != -1)  continue;

      prev_x[son_y[i]] = i;

      if(find(son_y[i]))  return true;

    }

  }

  return false;

}

LL KM(){

  ms(son_y, -1);  ms(y, 0);

  for(int i = 0; i < n; ++i){

    x[i] = 0;

    for(int j = 0; j < n; ++j)

      x[i] = max(x[i], w[i][j]);

  }

  bool flag;

  for(int i = 0; i < n; ++i){

    for(int j = 0; j < n; ++j){

      prev_x[j] = prev_y[j] = -1;

      slack[j] = LNF;

    }

    prev_x[i] = -2;

    if(find(i))  continue;

    flag = false;

    while(!flag){

      LL m = LNF;

      for(int j = 0; j < n; ++j)

        if(prev_y[j] == -1)  m = min(m, slack[j]);

      for(int j = 0; j < n; ++j){

        if(prev_x[j] != -1)  x[j] -= m;

        if(prev_y[j] != -1)  y[j] += m;

        else  slack[j] -= m;

      }

      for(int j = 0; j < n; ++j)  if(prev_y[j] == -1 && !slack[j]){

        prev_y[j] = par[j];

        if(son_y[j] == -1){

          adjust(j);

          flag = true;

          break;

        }

        prev_x[son_y[j]] = j;

        if(find(son_y[j])){

          flag = true;  break;

        }

      }

    }

  }

  LL ans = 0;

  for(int i = 0; i < n; ++i)  ans += w[son_y[i]][i];

  return ans;

}

int main(){

  int T;  cin >> T;

  for(int kase = 1; kase <= T; ++kase){

    scanf("%d", &n);

    for(int i = 0; i < n; ++i)

      for(int j = 0; j < n; w[i][j] = -w[i][j], ++j)

        scanf("%I64d", &w[i][j]);

    printf("Case #%d: %I64d\n", kase, -KM());

  }

  return 0;

}