角色扮演

题意：

有n种技能，每个技能有两个属性，分别对应着物理伤害，魔法伤害，现在有m个连招，如果连续两个技能都选择同一种属性就会得到一个属性加成，否则就会丢失一些伤害，现在需要使得伤害最大化，求这个最大的伤害。

题解：

因为两个属性物理或者魔法只能选择一种属性，那么很容易就想到最小割，那么再用总伤害减去最小割就是最大的技能伤害，但是对于伤害加成和伤害丢失怎么办呢？有一种建图方式。

注：S 到 p1的容量是AP（魔法）加成 + 伤害丢失 ， p2 到 T的容量是AD（物理）加成 + 伤害丢失。

代码:

#include <bits/stdc++.h>
using namespace std;
#define LL long long
const int M = 1e6 + 7;
const int N = 1e6 + 7;
const int INF = (1 << 31) - 1;
struct edge
{
	int v, c, f, nxt;
} e[M << 1];
LL sum;
int vis[N], S, T, ecnt, head[N], dep[N], cur[N], n, m;

void adde (int u, int v, int c)
{
	e[ecnt] = (edge) {v, c, 0, head[u]}, head[u] = ecnt++;
	e[ecnt] = (edge) {u, 0, 0, head[v]}, head[v] = ecnt++;
}

int BFS (int S)
{
	queue <int> q;
	q.push(S);
	memset (vis, 0, sizeof vis);
	vis[S] = 1, dep[S] = 0;
	while (!q.empty())
	{
		int u = q.front();
		q.pop();
		for (int it = head[u]; it != -1; it = e[it].nxt)
		{
			int v = e[it].v;
			if (!vis[v] && e[it].c > e[it].f)
			{
				vis[v] = 1;
				dep[v] = dep[u] + 1;
				q.push(v);
			}
		}
	}
	return vis[T];
}

LL DFS (int u, LL flow)
{
	if (u == T || flow == 0) return flow;
	LL tot = 0, F;
	for (int it = cur[u]; it != -1; it = e[it].nxt)
	{
		cur[u] = it;
		int v = e[it].v;
		if (dep[u] + 1 == dep[v] && (F = DFS(v, min(flow, (LL) e[it].c - e[it].f))) > 0)
		{
			e[it].f += F;
			e[it ^ 1].f -= F;
			tot += F;
			flow -= F;
			if (flow == 0) break;
		}
	}
	return tot;
}

LL dinic ()
{
	LL maxf = 0;
	while (BFS(S))
	{
		for (int i = 0; i <= T; ++ i) cur[i] = head[i];
		maxf += DFS(S, INF);
	}
	return maxf;
}

int main ()
{
	memset (head, -1, sizeof head);
	scanf ("%d%d", &n, &m);
	T = 2 * m + n + 1;
	for (int i = 1; i <= n; ++ i)
	{
		int P, D;
		scanf ("%d%d", &P, &D);
		sum += P + D;
		adde (S, i, P);
		adde (i, T, D);
	}
	int tmp = n + 1;
	for (int i = 1; i <= m; ++ i)
	{
		int u, v, P, D, z;
		scanf ("%d%d%d%d%d", &u, &v, &D, &P, &z);
		sum += P + D + z;
		adde (S, tmp, P + z);
		adde (tmp, u, INF);
		adde (tmp, v, INF);
		++tmp;
		adde (u, tmp, INF);
		adde (v, tmp, INF);
		adde (tmp, T, D + z);
		++tmp;
	}
	cout << sum - dinic() << endl;
	return 0;
}