題目鏈接: (bzoj) https://www.lydsy.com/JudgeOnline/problem.php?id=3894

(luogu) https://www.luogu.org/problemnew/show/P4313

題解:

做法很簡(jiǎn)單，就是最小割，\(S\)集屬于文科，\(T\)集屬于理科，對(duì)于每個(gè)點(diǎn)\(i\), 起點(diǎn)\(S\)向\(i\)連\(a_i\)(文科收益/理科代價(jià))，\(i\)向終點(diǎn)\(T\)連\(b_i\) (理科收益/文科代價(jià))，對(duì)于每一個(gè)點(diǎn)\(i\)再新建兩點(diǎn)\(i_a\)(同文點(diǎn))和\(i_b\)(同理點(diǎn))，\(S\)向\(i_a\)連邊\(aa_i\)(同文收益)，\(i_b\)向\(T\)連\(bb_i\)(同理收益)，中間對(duì)于\(i\)和\(i\)座位相連的每個(gè)點(diǎn)，從\(i_a\)向該點(diǎn)連邊，從該點(diǎn)向\(i_b\)連邊，邊權(quán)均為\(+\inf\).

我的錯(cuò)誤做法: 如果同文同理建成同一個(gè)點(diǎn)，和座位相連的每個(gè)點(diǎn)連雙向邊，那么這是錯(cuò)的，如果連單向邊也是錯(cuò)的。因?yàn)榻▋蓚€(gè)點(diǎn)實(shí)際上可以保證如果\(i_a\)屬于\(S\)集則它連向的人都選文，如果\(i_b\)屬于\(T\)集則連向它的人都選理，如果它們與\(S,T\)之間的邊都被割掉了，則它們對(duì)這些人沒(méi)有任何限制，這些人仍是獨(dú)立的。但如果同文同理建成同一個(gè)點(diǎn)連雙向邊，那么這些點(diǎn)之間構(gòu)成強(qiáng)連通分量，相當(dāng)于默認(rèn)所有人必須在同一集合，這是最離譜的做法我居然能想出來(lái)。如果連單向邊呢，比如從新點(diǎn)往這幾個(gè)人連邊，從\(S\)往新點(diǎn)連邊，從新點(diǎn)往\(T\)連邊，那么相當(dāng)于規(guī)定“如果新點(diǎn)屬于\(S\)則這些人全屬于\(S\), 如果新點(diǎn)屬于\(T\)則對(duì)這些人沒(méi)有要求”。總之，從\(i\)往\(j\)連邊\(\inf\)則相當(dāng)于如果\(i\in S\)則\(j\in S\), 但是如果\(i\in T\)則對(duì)\(j\)沒(méi)有要求；如果\(j\in T\)則\(i\in T\)，而如果\(j\in S\)則沒(méi)有要求對(duì)\(i\)沒(méi)有要求(這兩句話(huà)是等價(jià)的)。\(i\)和\(j\)之間連雙向\(\inf\)邊則相當(dāng)于強(qiáng)制兩點(diǎn)在同一集合中。

代碼

#include<cstdio> #include<cstdlib> #include<cstring> #include<algorithm> #include<cassert> using namespace std;const int INF = 1e8;namespace MaxFlow {const int N = 3e4+2;const int M = 14e4;struct Edge{int v,w,nxt,rev;} e[(M<<1)+3];int fe[N+3];int te[N+3];int que[N+3];int dep[N+3];int n,en,s,t;void addedge(int u,int v,int w){en++; e[en].v = v; e[en].w = w;e[en].nxt = fe[u]; fe[u] = en; e[en].rev = en+1;en++; e[en].v = u; e[en].w = 0;e[en].nxt = fe[v]; fe[v] = en; e[en].rev = en-1;}bool bfs(){for(int i=1; i<=n; i++) dep[i] = 0;int head = 1,tail = 1; que[tail] = s; dep[s] = 1;while(head<=tail){int u = que[head]; head++;for(int i=fe[u]; i; i=e[i].nxt){if(dep[e[i].v]==0 && e[i].w>0){dep[e[i].v] = dep[u]+1;tail++; que[tail] = e[i].v;}}}return dep[t]!=0; }int dfs(int u,int cur){if(u==t) {return cur;}int rst = cur;for(int i=te[u]; i; i=e[i].nxt){if(dep[e[i].v]==dep[u]+1 && e[i].w>0 && rst>0){int flow = dfs(e[i].v,min(rst,e[i].w));if(flow>0){rst -= flow; e[i].w -= flow; e[e[i].rev].w += flow;if(e[i].w>0) {te[u] = i;}if(rst==0) return cur;}}}if(cur==rst) dep[u] = 0;return cur-rst;}int dinic(int _n,int _s,int _t){int ret = 0;n = _n,s = _s,t = _t;while(bfs()){for(int i=1; i<=n; i++) te[i] = fe[i];ret += dfs(s,INF);}return ret;} } using MaxFlow::addedge; using MaxFlow::dinic;const int N = 100; int a[N+3][N+3],b[N+3][N+3],aa[N+3][N+3],bb[N+3][N+3]; int n,m;int getid(int x,int y) {return (x-1)*m+y+2;}int main() {scanf("%d%d",&n,&m); int ans = 0;for(int i=1; i<=n; i++){for(int j=1; j<=m; j++){scanf("%d",&a[i][j]); ans += a[i][j];}}for(int i=1; i<=n; i++){for(int j=1; j<=m; j++){scanf("%d",&b[i][j]); ans += b[i][j];}}for(int i=1; i<=n; i++){for(int j=1; j<=m; j++){scanf("%d",&aa[i][j]); ans += aa[i][j];}}for(int i=1; i<=n; i++){for(int j=1; j<=m; j++){scanf("%d",&bb[i][j]); ans += bb[i][j];}}for(int i=1; i<=n; i++){for(int j=1; j<=m; j++){int x = getid(i,j);addedge(1,x,a[i][j]);addedge(x,2,b[i][j]);addedge(1,x+n*m,aa[i][j]);addedge(x+n*m*2,2,bb[i][j]);addedge(x+n*m,x,INF);addedge(x,x+n*m*2,INF);if(i>1){addedge(x+n*m,getid(i-1,j),INF);addedge(getid(i-1,j),x+n*m*2,INF);}if(j>1){addedge(x+n*m,getid(i,j-1),INF);addedge(getid(i,j-1),x+n*m*2,INF);}if(i<n){addedge(x+n*m,getid(i+1,j),INF);addedge(getid(i+1,j),x+n*m*2,INF);}if(j<m){addedge(x+n*m,getid(i,j+1),INF);addedge(getid(i,j+1),x+n*m*2,INF);}}}int tmp = dinic(n*m*3+2,1,2);ans -= tmp;printf("%d\n",ans);return 0; }

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