题目描述

有 分配问题 费用流_#include

输入输出格式

输入格式：

文件的第 分配问题 费用流_#include_02

接下来的 分配问题 费用流_i++_03

输出格式：

两行分别输出最小总效益和最大总效益。

输入输出样例

输入样例#1：

复制

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

输出样例#1： 复制

5
14

说明

分配问题 费用流_#include_04

一个人只能修一个工件

分配问题 费用流_#include_05

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
//#include
//#pragma GCC optimize(2)
using namespace std;
#define maxn 20005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
#define mclr(x,a) memset((x),a,sizeof(x))
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-5
typedef pair pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair pii;

inline int rd() {
  int x = 0;
  char c = getchar();
  bool f = false;
  while (!isdigit(c)) {
    if (c == '-') f = true;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 1) + (x << 3) + (c ^ 48);
    c = getchar();
  }
  return f ? -x : x;
}


ll gcd(ll a, ll b) {
  return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }



/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
  if (!b) {
    x = 1; y = 0; return a;
  }
  ans = exgcd(b, a%b, x, y);
  ll t = x; x = y; y = t - a / b * y;
  return ans;
}
*/

bool vis[maxn];
int n, m, s, t;
int x, y, f, z;
int dis[maxn], pre[maxn], last[maxn], flow[maxn];
int maxflow, mincost;

struct node {
  int to, nxt, flow, dis;
}edge[maxn << 2];

int head[maxn], cnt;


void addedge(int from, int to, int flow, int dis) {
  edge[++cnt].to = to; edge[cnt].flow = flow; edge[cnt].dis = dis;
  edge[cnt].nxt = head[from]; head[from] = cnt;
}

bool spfa1(int s, int t) {
  memset(dis, 0x7f, sizeof(dis)); 
  memset(flow, 0x7f, sizeof(flow));
  ms(vis);queueq;
  q.push(s); vis[s] = 1; dis[s] = 0; pre[t] = -1;
  while (!q.empty()) {
    int now = q.front(); q.pop(); vis[now] = 0;
    for (int i = head[now]; i != -1; i = edge[i].nxt) {
      if (edge[i].flow > 0 && dis[edge[i].to] > dis[now] + edge[i].dis) {
        dis[edge[i].to] = edge[i].dis + dis[now];
        pre[edge[i].to] = now; last[edge[i].to] = i;
        flow[edge[i].to] = min(flow[now], edge[i].flow);
        if (!vis[edge[i].to]) {
          vis[edge[i].to] = 1; q.push(edge[i].to);
        }
      }
    }
  }
  return pre[t] != -1;
}
bool spfa2(int s, int t) {
  memset(dis, 0x7f, sizeof(dis));
  memset(flow, 0x7f, sizeof(flow));
  ms(vis); queueq;
  q.push(s); vis[s] = 1; dis[s] = 0; pre[t] = -1;
  while (!q.empty()) {
    int now = q.front(); q.pop(); vis[now] = 0;
    for (int i = head[now]; i != -1; i = edge[i].nxt) {
      if (edge[i].flow > 0 && dis[edge[i].to] > dis[now] + edge[i].dis) {
        dis[edge[i].to] = edge[i].dis + dis[now];
        pre[edge[i].to] = now; last[edge[i].to] = i;
        flow[edge[i].to] = min(flow[now], edge[i].flow);
        if (!vis[edge[i].to]) {
          vis[edge[i].to] = 1; q.push(edge[i].to);
        }
      }
    }
  }
  return pre[t] != -1;
}

void mincost_maxflow() {
  while (spfa1(s, t)) {
    int now = t;
    maxflow += flow[t]; mincost += flow[t] * dis[t];
    while (now != s) {
      edge[last[now]].flow -= flow[t];
      edge[last[now] ^ 1].flow += flow[t];
      now = pre[now];
    }
  }
}
void maxcost_maxflow() {
  while (spfa2(s, t)) {
    int now = t;
    maxflow += flow[t]; mincost += flow[t] * dis[t];
    while (now != s) {
      edge[last[now]].flow -= flow[t];
      edge[last[now] ^ 1].flow += flow[t];
      now = pre[now];
    }
  }
}

int C[104][104];

int main()
{
  //  ios::sync_with_stdio(0);
  n = rd(); mclr(head, -1); cnt = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++)C[i][j] = rd();
  }
  s = 0; t = 2 * n + 2;
  for (int i = 1; i <= n; i++) {
    addedge(s, i, 1, 0); addedge(i, s, 0, 0);
  }
  for (int i = 1; i <= n; i++) {
    addedge(i + n, t, 1, 0); addedge(t, i + n, 0, 0);
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      addedge(i, j + n, 1, C[i][j]); addedge(j + n, i, 0, -C[i][j]);
    }
  }
  int ans1 = 0, ans2 = 0;
  mincost_maxflow(); ans1 = mincost;
  mclr(head, -1); cnt = 1;
  ms(edge); ms(pre); ms(last); mincost = maxflow = 0;
  for (int i = 1; i <= n; i++) {
    addedge(s, i, 1, 0); addedge(i, s, 0, 0);
  }
  for (int i = 1; i <= n; i++) {
    addedge(i + n, t, 1, 0); addedge(t, i + n, 0, 0);
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      addedge(i, j + n, 1, -C[i][j]); addedge(j + n, i, 0, C[i][j]);
    }
  }
  maxcost_maxflow();
  ans2 = mincost;
  cout << ans1 << endl; cout << -ans2 << endl;
  return 0;
}

 

EPFL - Fighting