图论专题测试 【最短路】【网络流】解题报告

今天进行了一个测试。比上次简单些。。。

第一题

思路

找一个最短路的值，还要记录路径。SPFA再跑一遍找一下还有哪些路径。但是注意，会被卡的！加了优化还是只有60分。所以就看看代码吧。

代码

先是我的：

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<cmath>#include<vector>#include<queue>using namespace std;const int N=300+5;const int M=1e5;const int INF=1e5*N;int n,m,ans=1,flag[N],dis[N],pre[N],head[N],num;deque<int> q;struct edge{    int u,v,w,next;    edge(){next=-1;}}ed[M+5];void build(int u,int v,int w){    ed[++num].u=u;    ed[num].v=v;    ed[num].w=w;    ed[num].next=head[u];    head[u]=num;}int SPFA(int s,int e){    for (int i=1;i<=n;i++)    dis[i]=(i==s)?0:INF,flag[i]=(i==s)?1:0;    while(!q.empty())q.pop_front();    q.push_back(s);    while(!q.empty())    {        int u=q.front();q.pop_front();        flag[u]=0;        for (int i=head[u];i!=-1;i=ed[i].next)        {            int v=ed[i].v;            if (dis[v]>dis[u]+ed[i].w)            {                dis[v]=dis[u]+ed[i].w;                pre[v]=i;                if (!flag[v])                {                    flag[v]=1;                    if (!q.empty())                    {                        if (dis[v]>dis[q.front()])q.push_back(v);                        else q.push_front(v);                    }                    else q.push_back(v);                }            }        }    }    return dis[e];}int main(){    freopen("change.in","r",stdin);    freopen("change.out","w",stdout);    memset(head,-1,sizeof(head));    scanf("%d%d",&n,&m);    for (int i=1;i<=m;i++)    {        int u,v,w;        scanf("%d%d%d",&u,&v,&w);        build(u,v,w);    }    int minn=SPFA(1,n);    while(true)    {        for (int u=n;u!=1;u=ed[pre[u]].u)        ed[pre[u]].w=INF;        int _minn=SPFA(1,n);        if (minn==_minn)ans++;        else break;    }    printf("%d",ans);    return 0;}

然后：

#include<stdio.h>#include<cstring>#include<algorithm>#include<queue>#include<map>using namespace std;struct edge{    int v,last,mrk,u,w;}ed[400010];priority_queue < pair<int,int> ,vector< pair<int,int> > ,greater<pair<int,int> > > state;int head[100010],se[100010],dis[100010],frm[400010];int n,m,ans,bns,cns,num=0,INF=1e9+7;void add(int u,int v,int w){    num++;    ed[num].v=v;    ed[num].w=w;    ed[num].u=u;    ed[num].mrk=0;    ed[num].last=head[u];    head[u]=num;}int spfa(int flag){    memset(frm,0,sizeof(frm));    memset(se,0,sizeof(se));    while(!state.empty()) state.pop();    for(int i=1;i<=n;i++) dis[i]=INF;    state.push(make_pair(0,1));    se[1]=1,dis[1]=0;    while(!state.empty())    {        int u=state.top().second;        se[u]=0;        state.pop();        for(int i=head[u];i;i=ed[i].last)        if(ed[i].mrk==0)        {            int v=ed[i].v;            if(dis[v]>dis[u]+ed[i].w)            {                dis[v]=dis[u]+ed[i].w;                frm[v]=i;                if(!se[v])                {                    state.push(make_pair(dis[v],v));                    se[v]=1;                }            }        }    }    if(flag==0) cns=dis[n];    int p=frm[n];    while(p)    {        ed[p].mrk=1;        p=frm[ed[p].u];    }    return dis[n];}void sov(int flag){    int bns;    while(1)    {        bns=spfa(flag);        if(bns!=cns) break ;        flag++,ans++;    }}int main(){    freopen("change.in","r",stdin);    freopen("change.out","w",stdout);    scanf("%d%d",&n,&m);    for(int i=1;i<=m;i++)    {        int u,v,w;        scanf("%d%d%d",&u,&v,&w);        add(u,v,w);    }    sov(0);    printf("%d",ans);}/*Whoso pulleth out this sword from this stone and anvil is duly born King of all England*/

第二题

思路：

网络流。二分等级，把比这个mid小的加上。然后分奇偶（注意特判2）建图，边上记录火力值，然后二分图之间是INF，跑一边最大流最小割就好了。具体看代码。

代码

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<vector>#include<queue>using namespace std;const int INF=0x3f3f3f3f;const int N=5e5+5;struct edge{    int v,w,last;}ed[2*N];int head[N],dis[N],tim[N],pow[N],l[N],primes[4*N],isnot[4*N],x[4*N];int n,k,num=1,cns,S,T,big=0,ptot=0;int min(int x,int y){    if (x<=y) return x;    else return y;}int max(int x,int y){    if (x>=y) return x;    else return y;}void build(int u,int v,int w){    num++;    ed[num].v=v;    ed[num].w=w;    ed[num].last=head[u];    head[u]=num;}void getphi(int n){    isnot[1]=1;    for (int i=2;i<=n;i++)    if (!isnot[i])    {        primes[ptot++]=i;        for (int j=i+i;j<=n;j+=i)        isnot[j]=1;    }}int bfs(){    queue<int> state;    memset(dis,-1,sizeof(dis));    state.push(S);    dis[S]=0;    while(!state.empty())    {        int u=state.front();        state.pop();        for (int i=head[u];i;i=ed[i].last)        {            int v=ed[i].v;            if (dis[v]==-1&&ed[i].w>0)            {                dis[v]=dis[u]+1;                state.push(v);            }        }    }    if (dis[T]==-1) return 0;    return 1;}int dfs(int u,int low){    int a=0;    if (u==T) return low;    for (int i=head[u];i;i=ed[i].last)    {        int v=ed[i].v;        if (ed[i].w>0&&dis[v]==dis[u]+1)        {            int tmp=dfs(v,min(low-a,ed[i].w));            ed[i].w-=tmp,ed[i^1].w+=tmp;            a+=tmp;        }    }    return a;}int main(){    freopen("card.in","r",stdin);    freopen("card.out","w",stdout);    getphi(1000005);    scanf("%d%d",&n,&k);    for (int i=1;i<=n;i++)    {        scanf("%d%d%d",&pow[i],&tim[i],&l[i]);        big=max(big,l[i]);    }    int lf=1,rg=1e9+7,ans=-1;    while(lf<=rg)    {        memset(head,0,sizeof(head));        memset(x,0,sizeof(x));        int mid=(lf+rg)>>1,big=0,tar,t=0,tot=0;        num=1,cns=0;        for (int i=1;i<=n;i++)        if (l[i]<=mid&&pow[i]>big&&tim[i]==1) big=pow[i],tar=i;        for (int i=1;i<=n;i++)        if (l[i]<=mid&&(i==tar||tim[i]>1))        {            x[++t]=i;            tot+=pow[i];        }        S=0,T=t+2;//建立汇点源点         for (int i=1;i<=t;i++)        if (tim[x[i]]&1)        {            build(S,i,pow[x[i]]);            build(i,S,0);        }//奇数建边         else        {            build(i,T,pow[x[i]]);            build(T,i,0);        }//偶数建边         for (int i=1;i<=t;i++)        if (tim[x[i]]&1)        for (int j=1;j<=t;j++)        if ((~tim[x[j]]&1)&&!isnot[tim[x[i]]+tim[x[j]]])        {            build(i,j,INF);            build(j,i,0);        }//特判         while(bfs())        {            int bns=0;            while(bns=dfs(S,INF)) tot-=bns;        }        if (tot>=k) rg=(ans=mid)-1;        else lf=mid+1;    }    printf("%d",ans);    return 0;}

第三题

思路

SPFA找最短路。一样要优化！分成两种，一种是灵力，一种是时间。用tarjan找强连通分量（因为只有这样会产生“眩晕”），再最短路。
刚开始找强连通分量用的是SPFA。。。又被卡了。。。

代码

#include<queue>#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#include<cmath>using namespace std;const int N=1e5+5;int n,m,head[N],num,dfn[N],low[N],con[N],sta[N],flag[N],top,tim;long long dis[N];long long cost[N];struct edge{    int v,w;    int next;    edge(){next=-1;}}ed[N+5];void build(int u,int v,int w){    ed[++num].v=v;    ed[num].w=w;    ed[num].next=head[u];    head[u]=num;}void tarjan(int u){    dfn[u]=low[u]=++tim;    flag[sta[++top]=u]=1;    for (int i=head[u];i!=-1;i=ed[i].next)    {        int v=ed[i].v;        if (!dfn[v])        {            tarjan(v);            low[u]=min(low[u],low[v]);        }        else if (flag[v])low[u]=min(low[u],dfn[v]);    }    if (dfn[u]==low[u])    for (int now=0;now!= u;)    {        flag[now=sta[top--]]=0;        con[now]=u;    }}void SPFA(){    deque<int> q;    memset(flag,0,sizeof flag);    memset(dis,-1,sizeof dis);    memset(cost,-1,sizeof cost);    q.push_back(1);dis[1]=0,cost[1]=0;    while(!q.empty())    {        int u=q.front();        q.pop_front();        flag[u]=0;        for (int i=head[u];i!=-1;i=ed[i].next)        {            int v=ed[i].v;            int cc=con[v]==con[u]?cost[u]+1:cost[u];            if (cost[v]>cc||cost[v]==-1||(dis[v]>dis[u]+ed[i].w&&cost[v]==cc)||(dis[v]==-1&&cost[v]==cc))            {                cost[v]=cc;                dis[v]=dis[u]+ed[i].w;                if (!flag[v])                {                    flag[v]=1;                    if (!q.empty())                    {                        if (dis[v]>dis[q.front()])q.push_back(v);                        else q.push_front(v);                    }                    q.push_back(v);                }            }        }    }}int main(){    freopen("festival.in","r",stdin);    freopen("festival.out","w",stdout);    memset(head,-1,sizeof(head));    scanf("%d%d",&n,&m);    for (int i=1;i<=m;i++)    {        int u,v,w;        scanf("%d%d%d",&u,&v,&w);        build(v,u,w);    }    for (int i=1;i<=n;i++)    if (!con[i])tarjan(i);    SPFA();    for (int k=2;k<=n;k++)    {        if (dis[k]==dis[0]) printf("-1\n");        else printf("%I64d %I64d\n",cost[k],dis[k]);    }    return 0;}

以上