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

2018"百度之星"程序设计大赛 - 资格赛

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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;
}

原文地址：https://www.cnblogs.com/dwtfukgv/p/9482677.html