题目地址:HDU 4975
对这题简直无语。。。本来以为这题要用什么更先进的方法,结果还是老方法,这么卡时间真的好吗。。。。比赛的时候用了判环的方法,一直TLE。。后来换了矩阵DP的方式,加了加剪枝就过了。。无语了。。
代码如下:
#include <cstdio> #include <cstring> #include <algorithm> #include <iostream> #include <cstdio> #include <string> #include <cstring> #include <stdlib.h> #include <math.h> #include <ctype.h> #include <queue> #include <map> #include <set> #include <algorithm> using namespace std; const int INF=0x3f3f3f3f; int head[1010], source, sink, nv, cnt, vis[610][600], mp[600][600]; int cur[1010], num[1010], d[1010], pre[1010]; int a[600], b[600]; struct node { int u, v, cap, next; } edge[1000000]; void add(int u, int v, int cap) { edge[cnt].v=v; edge[cnt].cap=cap; edge[cnt].next=head[u]; head[u]=cnt++; edge[cnt].v=u; edge[cnt].cap=0; edge[cnt].next=head[v]; head[v]=cnt++; } void bfs() { memset(num,0,sizeof(num)); memset(d,-1,sizeof(d)); queue<int>q; q.push(sink); d[sink]=0; num[0]=1; while(!q.empty()) { int u=q.front(); q.pop(); for(int i=head[u]; i!=-1; i=edge[i].next) { int v=edge[i].v; if(d[v]==-1) { d[v]=d[u]+1; num[d[v]]++; q.push(v); } } } } int isap() { memcpy(cur,head,sizeof(cur)); int flow=0, u=pre[source]=source, i; bfs(); while(d[source]<nv) { if(u==sink) { int f=INF, pos; for(i=source; i!=sink; i=edge[cur[i]].v) { if(f>edge[cur[i]].cap) { f=edge[cur[i]].cap; pos=i; } } for(i=source; i!=sink; i=edge[cur[i]].v) { edge[cur[i]].cap-=f; edge[cur[i]^1].cap+=f; } flow+=f; u=pos; } for(i=cur[u]; i!=-1; i=edge[i].next) { if(d[edge[i].v]+1==d[u]&&edge[i].cap) break; } if(i!=-1) { cur[u]=i; pre[edge[i].v]=u; u=edge[i].v; } else { if(--num[d[u]]==0) break; int mind=nv; for(i=head[u]; i!=-1; i=edge[i].next) { if(mind>d[edge[i].v]&&edge[i].cap) { mind=d[edge[i].v]; cur[u]=i; } } d[u]=mind+1; num[d[u]]++; u=pre[u]; } } return flow; } bool panduan(int n,int m) { memset(vis,0,sizeof(vis)); int i, j, k; for(i=1;i<=n;i++) { if(a[i]==0||a[i]==9*m) continue ; for(j=1;j<=m;j++) { if(b[j]==0||b[j]==9*n) continue ; for(k=j+1;k<=m;k++) { int t1=0, t2=0; if(mp[i][j]&&mp[i][k]!=9) { if(vis[j][k]) return 1; t1=1; } if(mp[i][k]&&mp[i][j]!=9) { if(vis[k][j]) return 1; t2=1; } if(t1) vis[k][j]=1; if(t2) vis[j][k]=1; } } } return 0; } int read() { int x = 0; char ch = ' '; while(ch < '0' || ch > '9') ch = getchar(); while(ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar(); return x; } int main() { int t, i, j, n, m, sum1, sum2, num=0; t=read(); while(t--) { num++; n=read(); m=read(); sum1=sum2=0; memset(head,-1,sizeof(head)); cnt=0; source=0; sink=n+m+1; nv=sink+1; for(i=1; i<=n; i++) { a[i]=read(); sum1+=a[i]; } for(i=1; i<=m; i++) { b[i]=read(); sum2+=b[i]; } printf("Case #%d: ",num); if(sum1!=sum2) { puts("So naive!"); continue ; } for(i=1; i<=n; i++) { add(source,i,a[i]); for(j=1; j<=m; j++) { add(i,j+n,9); } } for(i=1; i<=m; i++) { add(i+n,sink,b[i]); } int ans=isap(); if(ans!=sum1) { puts("So naive!"); continue ; } for(i=1;i<=n;i++) { for(j=head[i];j!=-1;j=edge[j].next) { int v=edge[j].v; if(v>n&&v<=n+m) mp[i][v-n]=9-edge[j].cap; } } if(panduan(n,m)) { puts("So young!"); } else puts("So simple!"); } return 0; }
HDU 4975 (杭电多校 #10 1005题)A simple Gaussian elimination problem.(网络流之最大流),布布扣,bubuko.com
HDU 4975 (杭电多校 #10 1005题)A simple Gaussian elimination problem.(网络流之最大流)
原文:http://blog.csdn.net/scf0920/article/details/38740231