题意：略
唯一会做的...
一眼最小割
就是最大权闭合子图呀
\(s\rightarrow d_{positive} \rightarrow -d_{negtive} \rightarrow t\)
然后区间包含关系连inf
然后向t连花费
一开始看成\(mx^2 + cx\) x是选择种类数，直接吓哭了 平方怎么割啊我只会费用流
然后发现x是编号gg
然后建图注意编号最大是1000，tle了两次....
#include#include #include #include #include using namespace std;typedef long long ll;const int N = 2e4+5, M = 4e6+5, INF = 1e9;inline int read() { char c=getchar(); int x=0,f=1; while(c<'0' || c>'9') {if(c=='-')f=-1; c=getchar();} while(c>='0' && c<='9') {x=x*10+c-'0'; c=getchar();} return x*f;}int n, m, a[105], d[105][105], s, t, sum, mx;struct edge{int v, ne, c, f;} e[M];int cnt=1, h[N];inline void ins(int u, int v, int c) { //printf("ins (%d, %d) %d\n", u, v, c); e[++cnt] = (edge){v, h[u], c, 0}; h[u] = cnt; e[++cnt] = (edge){u, h[v], 0, 0}; h[v] = cnt;}namespace mf { int vis[N], d[N], q[N], head, tail; bool bfs() { memset(vis, 0, sizeof(vis)); head = tail = 1; q[tail++] = s; d[s] = 0; vis[s] = 1; while(head != tail) { int u = q[head++]; for(int i=h[u];i;i=e[i].ne) if(!vis[e[i].v] && e[i].c > e[i].f) { int v = e[i].v; d[v] = d[u]+1; vis[v] = 1; q[tail++] = v; if(v == t) return true; } } return false; } int cur[N]; int dfs(int u, int a) { if(u == t || a == 0) return a; int flow = 0, f; for(int &i=cur[u];i;i=e[i].ne) if(d[e[i].v] == d[u]+1 && (f = dfs(e[i].v, min(a, e[i].c - e[i].f))) > 0) { flow += f; e[i].f += f; e[i^1].f -= f; a -= f; if(a == 0) break; } if(a) d[u] = -1; return flow; } int dinic() { int flow = 0; while(bfs()) { for(int i=s; i<=t; i++) cur[i] = h[i]; flow += dfs(s, INF); } return flow; }}inline int id(int i, int j) {return i==j ? i : i*n+j;}int vis[N];void build() { int n2 = n*n, n3 = n+n2; s = 0; t = n + n2 + mx + 1; for(int i=1; i<=n; i++) { if(d[i][i] > 0) ins(s, i, d[i][i]); else ins(i, t, -d[i][i]); ins(i, t, a[i]); if(m == 0) continue; ins(i, n3 + a[i], INF); if(!vis[a[i]]) ins(n3 + a[i], t, a[i] * a[i]), vis[a[i]] = 1; } for(int i=n; i>=1; i--) for(int j=i+1; j<=n; j++) { int now = id(i, j); if(d[i][j] > 0) ins(s, now, d[i][j]); else {ins(now, t, -d[i][j]); continue;} int rr = 0; for(int l=i; l<=j; l++) { int _ = max(i, rr); for(int r= l==i ? j-1 : j; r>=_; r--) { ins(now, id(l, r), INF); if(d[l][r] > 0) { rr = max(rr, r); break; } } } }}int main() { freopen("in", "r", stdin); n=read(); m=read(); for(int i=1; i<=n; i++) a[i] = read(), mx = max(mx, a[i]); for(int i=1; i<=n; i++) for(int j=i; j<=n; j++) { d[i][j] = read(); if(d[i][j] > 0) sum += d[i][j]; } build(); int ans = mf::dinic(); //printf("ans %d\n", ans); printf("%d\n", sum - ans);}