POJ3422：Kaka's Matrix Travels——題解
阿新 • • 發佈：2017-12-02
string 當前 -a 我們 ron ont color ems namespace
http://poj.org/problem?id=3422
題目大意：
從左上角走到右下角，中途取數（數>=0），然後該點的數變為0，求走k的總價值和最大值。
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最大值？但是我們只會最小費用流啊……
但是數是>=0的啊，所以……
我們拆點，中間連一條容量為1費用為當前值負值的邊，再連一條容量為k-1費用0的邊。
這樣就一定會先走前一條邊啦！
然後向下向右連一條容量為k費用0的邊。
源點到左上角連一條容量為k-1費用0的邊。
右下角到匯點連一條容量為k-1費用0的邊。
#include<cstdio> #include<iostream> #include<queue> #include<cstring> #include<algorithm> #include<cctype> usingnamespace std; typedef long long ll; const int INF=1e9; const int N=50*50*2+10; inline int read(){ int X=0,w=0;char ch=0;#include<cstdio>
#include<iostream>
#include<queue>
#include<cstring>
#include<algorithm>
#include<cctype>
using namespace std;
typedef long long ll;const int INF=1e9;
const int N=50*50*2+10;
inline int read(){
int X=0,w=0;char ch=0;
while(!isdigit(ch)){w|=ch==‘-‘;ch=getchar();}
while(isdigit(ch))X=(X<<3)+(X<<1)+(ch^48),ch=getchar();
return w?-X:X;
}
struct node{
int nxt;
int to;
int w;
int b;
}edge[N*N*2];
int head[N],cnt=-1;
void add(int u,int v,int w,int b){
cnt++;
edge[cnt].to=v;
edge[cnt].w=w;
edge[cnt].b=b;
edge[cnt].nxt=head[u];
head[u]=cnt;
return;
}
int dis[N];
bool vis[N];
inline bool spfa(int s,int t,int n){
deque<int>q;
memset(vis,0,sizeof(vis));
for(int i=1;i<=n;i++)dis[i]=INF;
dis[t]=0;q.push_back(t);vis[t]=1;
while(!q.empty()){
int u=q.front();
q.pop_front();vis[u]=0;
for(int i=head[u];i!=-1;i=edge[i].nxt){
int v=edge[i].to;
int b=edge[i].b;
if(edge[i^1].w&&dis[v]>dis[u]-b){
dis[v]=dis[u]-b;
if(!vis[v]){
vis[v]=1;
if(!q.empty()&&dis[v]<dis[q.front()]){
q.push_front(v);
}else{
q.push_back(v);
}
}
}
}
}
return dis[s]<INF;
}
int ans=0;
int dfs(int u,int flow,int m){
if(u==m){
vis[m]=1;
return flow;
}
int res=0,delta;
vis[u]=1;
for(int e=head[u];e!=-1;e=edge[e].nxt){
int v=edge[e].to;
int b=edge[e].b;
if(!vis[v]&&edge[e].w&&dis[u]-b==dis[v]){
delta=dfs(v,min(edge[e].w,flow-res),m);
if(delta){
edge[e].w-=delta;
edge[e^1].w+=delta;
res+=delta;
ans+=delta*b;
if(res==flow)break;
}
}
}
return res;
}
inline int costflow(int S,int T,int n){
while(spfa(S,T,n)){
do{
memset(vis,0,sizeof(vis));
dfs(S,INF,T);
}while(vis[T]);
}
return ans;
}
int main(){
memset(head,-1,sizeof(head));
int n=read();
int k=read();
int S=2*n*n+1,T=S+1;
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++){
int a=read();
int pos=(i-1)*n+j;
add(pos,pos+n*n,1,-a);
add(pos+n*n,pos,0,a);
add(pos,pos+n*n,k-1,0);
add(pos+n*n,pos,0,0);
pos+=n*n;
if(i+1<=n){
int ppos=i*n+j;
add(pos,ppos,k,0);
add(ppos,pos,0,0);
}
if(j+1<=n){
int ppos=(i-1)*n+j+1;
add(pos,ppos,k,0);
add(ppos,pos,0,0);
}
}
}
add(S,1,k,0);
add(1,S,0,0);
add(2*n*n,T,k,0);
add(T,2*n*n,0,0);
printf("%d\n",-costflow(S,T,T));
return 0;
}
while(!isdigit(ch)){w|=ch==‘-‘;ch=getchar();} while(isdigit(ch))X=(X<<3)+(X<<1)+(ch^48),ch=getchar(); return w?-X:X; } struct node{ int nxt; int to; int w; int b; }edge[N*N*2]; int head[N],cnt=-1; void add(int u,int v,int w,int b){ cnt++; edge[cnt].to=v; edge[cnt].w=w; edge[cnt].b=b; edge[cnt].nxt=head[u]; head[u]=cnt; return; } int dis[N]; bool vis[N]; inline bool spfa(int s,int t,int n){ deque<int>q; memset(vis,0,sizeof(vis)); for(int i=1;i<=n;i++)dis[i]=INF; dis[t]=0;q.push_back(t);vis[t]=1; while(!q.empty()){ int u=q.front(); q.pop_front();vis[u]=0; for(int i=head[u];i!=-1;i=edge[i].nxt){ int v=edge[i].to; int b=edge[i].b; if(edge[i^1].w&&dis[v]>dis[u]-b){ dis[v]=dis[u]-b; if(!vis[v]){ vis[v]=1; if(!q.empty()&&dis[v]<dis[q.front()]){ q.push_front(v); }else{ q.push_back(v); } } } } } return dis[s]<INF; } int ans=0; int dfs(int u,int flow,int m){ if(u==m){ vis[m]=1; return flow; } int res=0,delta; vis[u]=1; for(int e=head[u];e!=-1;e=edge[e].nxt){ int v=edge[e].to; int b=edge[e].b; if(!vis[v]&&edge[e].w&&dis[u]-b==dis[v]){ delta=dfs(v,min(edge[e].w,flow-res),m); if(delta){ edge[e].w-=delta; edge[e^1].w+=delta; res+=delta; ans+=delta*b; if(res==flow)break; } } } return res; } inline int costflow(int S,int T,int n){ while(spfa(S,T,n)){ do{ memset(vis,0,sizeof(vis)); dfs(S,INF,T); }while(vis[T]); } return ans; } int main(){ memset(head,-1,sizeof(head)); int n=read(); int k=read(); int S=2*n*n+1,T=S+1; for(int i=1;i<=n;i++){ for(int j=1;j<=n;j++){ int a=read(); int pos=(i-1)*n+j; add(pos,pos+n*n,1,-a); add(pos+n*n,pos,0,a); add(pos,pos+n*n,k-1,0); add(pos+n*n,pos,0,0); pos+=n*n; if(i+1<=n){ int ppos=i*n+j; add(pos,ppos,k,0); add(ppos,pos,0,0); } if(j+1<=n){ int ppos=(i-1)*n+j+1; add(pos,ppos,k,0); add(ppos,pos,0,0); } } } add(S,1,k,0); add(1,S,0,0); add(2*n*n,T,k,0); add(T,2*n*n,0,0); printf("%d\n",-costflow(S,T,T)); return 0; }
POJ3422：Kaka's Matrix Travels——題解