HDU 4975 (杭电多校 #10 1005题）A simple Gaussian elimination problem.（网络流之最大流）

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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;}

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