POJ3422: Kaka's Matrix Travel 题解

这一题乍一看和费用流模型不太一样，因为一个点不论经过多少次，都只加一次权值

于是我们可以这样做：首先拆点，将每个格子拆成in和out

对于每个格子，向下方和右方（如果存在）连一条流量为INF，费用为0的边，从out连到in

对于每个格子拆成的两个点，从in向out连一条流量为1，费用为-a[i][j]的边，再连一条流量为INF，费用为0的边

这样走第一次的时候一定会优先选择费用较小的那条边

但是他的流量只有1，所以流过一次后，再流这个格子就只能走费用为0的边，相当于一个格子的数字只计算了一次

注意刚开始有负权边，要先跑一趟spfa

#include <cstdio>#include <iostream>#include <cstring>#include <string>#include <cmath>#include <algorithm>#include <cstdlib>#include <utility>#include <map>#include <stack>#include <set>#include <vector>#include <queue>#include <deque>#define x first#define y second#define mp make_pair#define pb push_back#define LL long long#define Pair pair<int,int>#define LOWBIT(x) x & (-x)using namespace std;const int MOD=1e9+7;const int INF=1e9;const int magic=348;int t,tot=1,head[100048],nxt[200048],to[200048],f[200048],w[200048];inline void addedge(int s,int t,int cap,int cost){to[++tot]=t;nxt[tot]=head[s];head[s]=tot;f[tot]=cap;w[tot]=cost;to[++tot]=s;nxt[tot]=head[t];head[t]=tot;f[tot]=0;w[tot]=-cost;}int n,k;int a[68][68];inline int getID(int x,int y,bool type){if (!type) return (x-1)*n+y; else return n*n+(x-1)*n+y;}int start,end;int dist[100048],h[100048];priority_queue<Pair> q;queue<int> qq;bool inq[100048];int prevv[100048],preve[100048];void spfa(){int i,x,y;for (i=start;i<=end;i++) dist[i]=INF;dist[start]=0;for (i=start;i<=end;i++) inq[i]=false;inq[start]=true;qq.push(start);while (!qq.empty()){x=qq.front();qq.pop();inq[x]=false;for (i=head[x];i;i=nxt[i]){y=to[i];if (f[i] && dist[y]>dist[x]+w[i]+h[x]-h[y]){dist[y]=dist[x]+w[i]+h[x]-h[y];prevv[y]=x;preve[y]=i;if (!inq[y]) qq.push(y);}}}}void dijkstra(){int i,x,y,dd;for (i=start;i<=end;i++) dist[i]=INF;dist[start]=0;q.push(mp(0,start));while (!q.empty()){x=q.top().y;dd=-q.top().x;q.pop();if (dd>dist[x]) continue;for (i=head[x];i;i=nxt[i]){y=to[i];if (f[i] && dist[y]>dist[x]+w[i]+h[x]-h[y]){dist[y]=dist[x]+w[i]+h[x]-h[y];prevv[y]=x;preve[y]=i;q.push(mp(-dist[y],y));}}}}int min_cost_flow(){int i,res,minf,u;for (i=start;i<=end;i++) h[i]+=dist[i];minf=INF;for (u=end;u!=start;u=prevv[u])minf=min(minf,f[preve[u]]);res=minf*h[end];for (u=end;u!=start;u=prevv[u]){f[preve[u]]-=minf;f[preve[u]^1]+=minf;}return res;}int main (){int i,j;scanf("%d%d",&n,&k);for (i=1;i<=n;i++)for (j=1;j<=n;j++)scanf("%d",&a[i][j]);for (i=1;i<=n;i++)for (j=1;j<=n;j++){addedge(getID(i,j,false),getID(i,j,true),1,-a[i][j]);addedge(getID(i,j,false),getID(i,j,true),INF,0);}for (i=1;i<=n;i++)for (j=1;j<=n;j++){if (i!=n) addedge(getID(i,j,true),getID(i+1,j,false),INF,0);if (j!=n) addedge(getID(i,j,true),getID(i,j+1,false),INF,0);}start=getID(1,1,false);end=getID(n,n,true);int ans=0;for (i=1;i<=k;i++){if (i==1) spfa(); else dijkstra();ans-=min_cost_flow();}printf("%d\n",ans);return 0;}