深海机器人问题
题解:
最大费用最大流。
建图很简单,就是将机器人作为流,进入就从$S$向内流,出来就从图向$T$流。
代码:
#include<queue> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; #define N 300 #define ll long long const int inf = 0x3f3f3f3f; const ll Inf = 0x3f3f3f3f3f3f3f3fll; inline int rd() { int f=1,c=0;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){c=10*c+ch-'0';ch=getchar();} return f*c; } int a,b,P,Q,S,T,hed[N],cnt=-1; int _id(int x,int y){return x*(Q+1)+y;} struct EG { int to,nxt; ll w,c; }e[10*N]; void ae(int f,int t,ll w,ll c) { e[++cnt].to = t; e[cnt].nxt = hed[f]; e[cnt].w = w; e[cnt].c = c; hed[f] = cnt; } queue<int>q; ll dis[N],fl[N]; int pre[N],fa[N]; bool vis[N]; bool spfa() { memset(dis,0x3f,sizeof(dis)); dis[S] = 0,fl[S] = Inf,vis[S] = 1; q.push(S); while(!q.empty()) { int u = q.front(); q.pop(); for(int j=hed[u];~j;j=e[j].nxt) { int to = e[j].to; if(e[j].w&&dis[to]>dis[u]+e[j].c) { dis[to] = dis[u]+e[j].c; fl[to] = min(fl[u],e[j].w); pre[to] = j,fa[to] = u; if(!vis[to]) { vis[to] = 1; q.push(to); } } } vis[u] = 0; } return dis[T]!=Inf; } ll mcmf() { ll ret = 0; while(spfa()) { ret+=fl[T]*dis[T]; int u = T; while(u!=S) { e[pre[u]].w-=fl[T]; e[pre[u]^1].w+=fl[T]; u=fa[u]; } } return ret; } int main() { a = rd(),b = rd(),P = rd(),Q = rd(); memset(hed,-1,sizeof(hed)); for(int i=0;i<=P;i++) for(int c,j=0;j<Q;j++) { int f = _id(i,j); int t = _id(i,j+1); c = rd(); ae(f,t,1,-c); ae(t,f,0,c); ae(f,t,Inf,0); ae(t,f,0,0); } for(int i=0;i<=Q;i++) for(int c,j=0;j<P;j++) { int f = _id(j,i); int t = _id(j+1,i); c = rd(); ae(f,t,1,-c); ae(t,f,0,c); ae(f,t,Inf,0); ae(t,f,0,0); } S = _id(P,Q)+1,T = S+1; for(int w,x,y,i=1;i<=a;i++) { w = rd(),x = rd(),y = rd(); int t = _id(x,y); ae(S,t,w,0); ae(t,S,0,0); } for(int w,x,y,i=1;i<=b;i++) { w = rd(),x = rd(),y = rd(); int f = _id(x,y); ae(f,T,w,0); ae(T,f,0,0); } printf("%lld\n",-mcmf()); return 0; }