Blog of RuSun

\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

P4210 土地划分

P4210 土地划分

经典的建图模型,考虑如何处理权值减少,实际上,只需要默认减少该权值,再在增加的权值上增加即可。

查看代码
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#include <cstdio>
#include <queue>
#include <cstring>
using namespace std;
const int N = 1e5 + 10, M = 6e5 + 10, inf = 2147483647;
int n, m, tot, sum, d[N], cur[N];
int idx = -1, hd[N], nxt[M], edg[M], wt[M];
bool bfs()
{
memset(d, -1, sizeof d);
d[1] = 0;
cur[1] = hd[1];
queue<int> q;
q.push(1);
while (!q.empty())
{
int t = q.front();
q.pop();
for (int i = hd[t]; ~i; i = nxt[i])
if (d[edg[i]] == -1 && wt[i])
{
d[edg[i]] = d[t] + 1;
cur[edg[i]] = hd[edg[i]];
if (edg[i] == n)
return true;
q.push(edg[i]);
}
}
return false;
}
int exploit(int x, int limit)
{
if (x == n)
return limit;
int res = 0;
for (int i = cur[x]; ~i && res < limit; i = nxt[i])
{
cur[x] = i;
if (d[edg[i]] == d[x] + 1 && wt[i])
{
int t = exploit(edg[i], min(limit - res, wt[i]));
if (!t)
d[edg[i]] = -1;
wt[i] -= t;
wt[i ^ 1] += t;
res += t;
}
}
return res;
}
int dinic()
{
int res = 0, flow;
while (bfs())
while (flow = exploit(1, inf))
res += flow;
return res;
}
void add(int a, int b, int c)
{
nxt[++idx] = hd[a];
hd[a] = idx;
edg[idx] = b;
wt[idx] = c;
}
int main()
{
scanf("%d%d", &n, &m);
tot = n;
memset(hd, -1, sizeof hd);
for (int i = 2, a; i < n; i++)
{
scanf("%d", &a);
add(1, i, a);
add(i, 1, 0);
sum += a;
}
for (int i = 2, a; i < n; i++)
{
scanf("%d", &a);
add(i, n, a);
add(n, i, 0);
sum += a;
}
for (int x, y, a, b, c; m; m--)
{
scanf("%d%d%d%d%d", &x, &y, &a, &b, &c);
if (x > y)
swap(x, y);
if (x == 1 && y == n)
{
sum -= c;
continue;
}
if (x == 1)
{
sum += a;
add(1, y, a + c);
add(y, 1, 0);
continue;
}
if (y == n)
{
sum += b;
add(x, n, b + c);
add(n, x, 0);
continue;
}
sum += a + b + c;
add(1, ++tot, a + c);
add(tot, 1, 0);
add(tot, x, inf);
add(x, tot, 0);
add(tot, y, inf);
add(y, tot, 0);
add(++tot, n, b + c);
add(n, tot, 0);
add(x, tot, inf);
add(tot, x, 0);
add(y, tot, inf);
add(tot, y, 0);
}
printf("%d", sum - dinic());
return 0;
}