POJ 3498 March of the Penguins

 

一看题就知道怎么做了。。。无奈代码能力不够，各种错误，各种WA，一开始邻接表写的SAP还TLE，不明白为什么

 

最后根据前人的经验修改后总算过了。。。

 

需要注意的是，每块冰可以承受的跳跃次数是点容量，需要拆点转换成边容量，具体做法是第i个点拆成i号和i+n号，i指向i+n,边上的容量为mi，i号点只能进，i+n号点只能出

 

然后枚举汇点就行

 

AC代码：（很慢，看着别人都几百MS，惭愧。。。）

#include<iostream>#include<cmath>#include<queue>using namespace std;#define min(x,y) (x)<(y)?(x):(y)#define inf 999999#define MAX 205struct nodee{double x;double y;int number;int m;}po[MAX];int n,S,T,max_flow,e,vn;int dis[MAX],num[MAX],pre[MAX],ans[MAX];int r[MAX][MAX];double maxDistance,d[MAX][MAX];double getDistance(double x1,double y1,double x2,double y2){double xx=x1-x2,yy=y1-y2;return sqrt(xx*xx+yy*yy);}void construct(int t){int i,j;for(i=1;i<=n;i++){r[i][i+n]=po[i].m;r[S][i]=po[i].number;}for(i=1;i<=n;i++)for(j=1;j<=n;j++){if(i==j)continue;if(d[i][j]<=maxDistance){r[i+n][j]=inf;}}/*for(i=1;i<=n;i++){if(d[i][t]<=maxDistance)r[i+n][T]=inf;}*/r[t][t+n]=inf;}void relabel(int k){int mm=vn,i;for(i=1;i<=vn;i++){if(r[k][i]>0&&dis[i]<mm)mm=dis[i];}dis[k]=mm+1;}void change(){int i,nf=inf;for(i=T;i!=S;i=pre[i]){if(r[pre[i]][i]<nf)nf=r[pre[i]][i];}for(i=T;i!=S;i=pre[i]){r[pre[i]][i]-=nf;r[i][pre[i]]+=nf;}max_flow+=nf;}void sap(){int i,j;for(i=0;i<=vn;i++){dis[i]=0;num[i]=0;}dis[0]=vn;i=S;while(dis[S]<vn){for(j=1;j<=vn;j++){if(r[i][j]>0&&dis[i]==dis[j]+1)break;}if(j>vn){num[dis[i]]--;if(num[dis[i]]==0)break;relabel(i);num[dis[i]]++;if(i!=S)i=pre[i];}else{pre[j]=i;i=j;if(i==T){change();i=S;}}}}int main(){int test,i,j,sum,cnt;cin>>test;while(test--){scanf("%d %lf",&n,&maxDistance);memset(d,0,sizeof(d));S=2*n+1;//T=2*n+2;vn=2*n+1;sum=0;for(i=1;i<=n;i++){scanf("%lf %lf %d %d",&po[i].x,&po[i].y,&po[i].number,&po[i].m);sum+=po[i].number;}for(i=1;i<=n;i++){for(j=i+1;j<=n;j++){d[i][j]=d[j][i]=getDistance(po[i].x,po[i].y,po[j].x,po[j].y);}}cnt=0;for(i=1;i<=n;i++){memset(r,0,sizeof(r));T=i+n;construct(i);max_flow=0;sap();if(max_flow==sum){ans[cnt++]=i-1;}}if(cnt==0)printf("-1/n");else{printf("%d",ans[0]);for(i=1;i<cnt;i++)printf("% d",ans[i]);printf("/n");}}system("pause");return 0;}