bzoj4873
http://www.lydsy.com/JudgeOnline/problem.php?id=4873
最大权闭合子图。。。
建图:
1.d[i][j]:i->j区间的费用,d[i][j] > 0 ins(S,id(i,j),d[i][j]) 否则ins(id(i,j),T,-d[i][j]) 套路
2.对于寿司怎么搞,m=1,ins(种类,T,a[i]*a[i]),ins(寿司,种类,inf):必须割掉初始的费用,ins(寿司,T,a[i]),ins(区间,寿司, inf):每个区间割掉需要寿司的花费
3.ins(id(i,j),id(i+1,j),inf),ins(id(i,j),id(i,j-1),inf):选了大的区间的必须选小的区间
但是上面有一步可以改进,就是2的最后。因为选了大的一定会选小的,那么我们只用将[i,i]这个区间向寿司连边就行了。
记住a[i]有1000,并且汇点不要取太小。。。
#include<bits/stdc++.h> using namespace std; const int N = 100010, inf = 1 << 29; struct edge { int nxt, to, f; } e[N * 2]; int dis[N], used[N], head[N], q[N], iter[N], d[110][110], a[N]; int n, m, sum, T = 0, num = 0, cnt = 1; namespace maxflow { void link(int u, int v, int f) { e[++cnt].nxt = head[u]; head[u] = cnt; e[cnt].to = v; e[cnt].f = f; } void ins(int u, int v, int f) { link(u, v, f); link(v, u, 0); } bool bfs() { int l = 1, r = 0; q[++r] = 0; memset(dis, 0, sizeof(dis)); dis[0] = 1; while(l <= r) { int u = q[l++]; for(int i = head[u]; i; i = e[i].nxt) if(!dis[e[i].to] && e[i].f) { dis[e[i].to] = dis[u] + 1; q[++r] = e[i].to; } } return dis[T] > 0; } int dfs(int u, int delta) { if(u == T) return delta; int ret = 0; for(int &i = iter[u]; i && delta; i = e[i].nxt) if(e[i].f && dis[e[i].to] == dis[u] + 1) { int x = dfs(e[i].to, min(delta, e[i].f)); e[i].f -= x; e[i ^ 1].f += x; ret += x; delta -= x; } return ret; } int id(int i, int j) { return (i - 1) * n + j; } void build() { //每个编号和T连边,每个寿司和对应编号连边 int D = n * n; T = N - 2; for(int i = 1; i <= n; ++i) { // i + D:寿司 a[i] + 2 * D: 种类 id(i, i): 区间 ins(i + D, T, a[i]); //每个寿司 if(d[i][i] < 0) ins(id(i, i), T, -d[i][i]); else ins(0, id(i, i), d[i][i]); ins(id(i, i), i + D, inf); if(!m) continue; if(!used[a[i]]) { used[a[i]] = 1; ins(a[i] + 2 * D, T, a[i] * a[i]); } ins(i + D, a[i] + 2 * D, inf); } for(int i = 1; i <= n; ++i) for(int j = i + 1; j <= n; ++j) { if(d[i][j] < 0) ins(id(i, j), T, -d[i][j]); else ins(0, id(i, j), d[i][j]); if(i < n) ins(id(i, j), id(i + 1, j), inf); if(j > 1) ins(id(i, j), id(i, j - 1), inf); } } int dinic() { int ret = 0; while(bfs()) { for(int i = 0; i <= T; ++i) iter[i] = head[i]; ret += dfs(0, inf); } return ret; } } using namespace maxflow; int main() { scanf("%d%d", &n, &m); for(int i = 1; i <= n; ++i) scanf("%d", &a[i]); for(int i = 1; i <= n; ++i) for(int j = i; j <= n; ++j) { scanf("%d", &d[i][j]); if(d[i][j] > 0) sum += d[i][j]; } build(); sum -= dinic(); printf("%d\n", sum); return 0; }