版权声明:本文为博主原创文章,未经博主允许不得转载。 https://blog.csdn.net/qq1059752567/article/details/78128297
思路:拆点,每个点拆成两个,限流的不用多说了吧。然后,关键在于某点A和B有边,但是实际结果只能二选一,怎么办呢?弄一个辅助点C加边方式(A’,C,1,-SC[A][B]), (B',C,1,-SC[B][A]), (C,C', 1, 0), (C',T, INF, 0); //(U,V,CAP,COST)
代码:
#include <stdio.h>
#include <string.h>
#include <vector>
#include <string>
#include <cmath>
#include <ctime>
#include <queue>
#include <stack>
#include <set>
#include <iostream>
#include <map>
#include <algorithm>
const double PI = acos(-1.0);
const double EPS = 1e-9;
using namespace std;
const int MAXN = 50000;
const int MAXM = 500000;
const int INF = 0x3f3f3f3f;
struct Edge
{
int to, next, cap, flow, cost;
}edge[MAXM];
int head[MAXN], tol;
int pre[MAXN], dis[MAXN];
bool vis[MAXN];
int N;//节点总个数,节点编号从0~N-1
void init()
{
tol = 0;
memset(head, -1, sizeof(head));
}
void addedge(int u, int v, int cap, int cost)
{
edge[tol].to = v;
edge[tol].cap = cap;
edge[tol].cost = cost;
edge[tol].flow = 0;
edge[tol].next = head[u];
head[u] = tol++;
edge[tol].to = u;
edge[tol].cap = 0;
edge[tol].cost = -cost;
edge[tol].flow = 0;
edge[tol].next = head[v];
head[v] = tol++;
}
bool spfa(int s, int t)
{
queue<int>q;
for (int i = 0; i < N; i++)
{
dis[i] = INF;
vis[i] = false;
pre[i] = -1;
}
dis[s] = 0;
vis[s] = true;
q.push(s);
while (!q.empty())
{
int u = q.front();
q.pop();
vis[u] = false;
for (int i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i].to;
if (edge[i].cap > edge[i].flow &&
dis[v] > dis[u] + edge[i].cost)
{
dis[v] = dis[u] + edge[i].cost;
pre[v] = i;
if (!vis[v])
{
vis[v] = true;
q.push(v);
}
}
}
}
if (pre[t] == -1)return false;
else return true;
}
//返回的是最大流,cost存的是最小费用
int minCostMaxflow(int s, int t, int &cost)
{
int flow = 0;
cost = 0;
while (spfa(s, t))
{
int Min = INF;
for (int i = pre[t]; i != -1; i = pre[edge[i ^ 1].to])
{
if (Min > edge[i].cap - edge[i].flow)
Min = edge[i].cap - edge[i].flow;
}
for (int i = pre[t]; i != -1; i = pre[edge[i ^ 1].to])
{
edge[i].flow += Min;
edge[i ^ 1].flow -= Min;
cost += edge[i].cost * Min;
}
flow += Min;
}
return flow;
}
bool flag[105][105];
int sc[105][105];
int n, st, ed;
void add(int i, int j)
{
addedge(i+n, N, 1, -sc[i][j]);
addedge(j+n, N, 1, -sc[j][i]);
N++;
addedge(N - 1, N, 1, 0);
addedge(N, ed, INF, 0);
N++;
}
int main()
{
int a[105];
while (scanf("%d", &n) != EOF)
{
ed = 2 * n + 1;
N = 2 * n + 2;
memset(flag, 0, sizeof(flag));
for (int i = 1; i <= n; i++)
scanf("%d", &a[i]);
int m;
scanf("%d", &m);
for (int i = 1; i <= m; i++)
{
int x, y;
scanf("%d%d", &x, &y);
a[x]--;
flag[x][y] = true;
}
int sum = 0;
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= n; j++)
{
scanf("%d", &sc[i][j]);
if (flag[i][j])
{
sum += sc[i][j];
sc[i][j] = -1;
}
if (flag[j][i])
sc[i][j] = -1;
}
}
init();
for (int i = 1; i <= n; i++)
{
addedge(st, i, INF, 0);
addedge(i, i + n, a[i], 0);
for (int j = i+1; j <= n; j++)
{
if (sc[i][j] != -1)
{
add(i, j);
}
}
}
int ans = 0;
minCostMaxflow(st, ed, ans);
cout << sum-ans << endl;
}
return 0;
}