图论

floyd
dijstra
spfa
bzoj1001

浅析最大最小定理在信息学竞赛中的应用

#include<iostream>#include<algorithm>#include<cstdio>#include<cstring>#include<cmath>#define MAXN 2000001#define INF   (~0U>>1)#define F(i,j,n)   for(int i=j;i<=n;i++)#include<queue>#include<vector>using namespace std;vector<int>edge[MAXN],weight[MAXN];bool vst[MAXN];int n,m,x,t,dis[MAXN];struct Heapnode{    int dist,num;    bool operator <(const Heapnode & x)const{        return dist>x.dist;    }};void solve(){    int x=max(n,m);    int p,minn=INF;    F(i,1,x-1)    {        scanf("%d",&p);        minn=min(p,minn);    }    if(minn==INF) minn=0;    printf("%d\n",minn);    return ;}priority_queue<Heapnode> Q;int main(){    scanf("%d%d",&n,&m);    if(n==1||m==1)    {        solve();        return 0;    }    t=(n-1)*(m-1)*2+1;    F(i,1,n)        F(j,1,m-1)        {            scanf("%d",&x);            if(i==1)            {                edge[t].push_back(2*j);                weight[t].push_back(x);                edge[2*j].push_back(t);                weight[2*j].push_back(x);            }            if(i==n)            {                int p=2*(n-2)*(m-1)+2*j-1;                edge[0].push_back(p);                weight[0].push_back(x);                edge[p].push_back(0);                weight[p].push_back(x);            }            if(i<n&&i>1)            {                int p=2*(m-1)*(i-1)+2*j;                int q=2*(m-1)*(i-2)+2*j-1;                edge[p].push_back(q);                weight[p].push_back(x);                edge[q].push_back(p);                weight[q].push_back(x);            }        }        F(i,1,n-1)            F(j,1,m)            {                scanf("%d",&x);                if(j==1)                {                    int p=2*(m-1)*(i-1)+1;                    edge[0].push_back(p);                    weight[0].push_back(x);                    edge[p].push_back(0);                    weight[p].push_back(x);                }                if(j==m)                {                    int p=2*(m-1)*i;                    edge[t].push_back(p);                    weight[t].push_back(x);                    edge[p].push_back(t);                    weight[p].push_back(x);                }                if(j>1&&j<m)                {                    int p=2*(m-1)*(i-1)+2*(j-1);                    edge[p].push_back(p+1);                    weight[p].push_back(x);                    edge[p+1].push_back(p);                    weight[p+1].push_back(x);                }            }    F(i,1,n-1)        F(j,1,m-1)        {            scanf("%d",&x);            int p=2*(m-1)*(i-1)+2*j-1;            edge[p].push_back(p+1);            weight[p].push_back(x);            edge[p+1].push_back(p);            weight[p+1].push_back(x);        }    F(i,1,t) dis[i]=INF;    dis[0]=0;    memset(vst,false,sizeof(vst));    Q.push((Heapnode){0,0});    while(!Q.empty())    {        Heapnode x=Q.top();        Q.pop();        if(vst[x.num]) continue;        vst[x.num]=true;        for(int i=0;i<edge[x.num].size();i++)        {            if(dis[edge[x.num][i]]>dis[x.num]+weight[x.num][i])            {                dis[edge[x.num][i]]=dis[x.num]+weight[x.num][i];                Q.push((Heapnode){dis[edge[x.num][i]],edge[x.num][i]});            }        }    }    printf("%d\n",dis[t]);    return 0;}

bzoj1726

#include<iostream>#include<algorithm>#include<cstdio>#include<cstring>#include<cmath>#include<queue>#define F(i,j,n) for(int i=j;i<=n;i++)#define inf  (~0U>>1)using namespace std;inline int read(){    int num=0,f=1;    char ch=getchar();    while(ch<'0'||ch>'9'){        if(ch=='-') f=-1;        ch=getchar();    }    while(ch<='9'&&ch>='0'){        num=num*10+ch-'0';        ch=getchar();    }    return num;}int m,n,d,head[5005],cnt;int dis[5005][2];int tmp,ans=inf;int a,b,x;bool v[5005];struct edge{    int f,t,l;  }e[200005];queue<int> Q;void add(int f,int t,int l){    e[++cnt].f=head[f];    e[cnt].t=t;    e[cnt].l=l;    head[f]=cnt;}void spfa(int start,int num){    memset(v,0,sizeof(v));    dis[start][num]=0;    Q.push(start);    v[start]=1;    while(!Q.empty()){        int y=Q.front();        Q.pop();        v[y]=0;        for(int i=head[y];i;i=e[i].f){            if(dis[e[i].t][num]>dis[y][num]+e[i].l)            {                dis[e[i].t][num]=dis[y][num]+e[i].l;                if(!v[e[i].t])                {                    Q.push(e[i].t);                    v[e[i].t]=1;                }            }        }       }}int main(){    memset(dis,0x7f,sizeof(dis));    n=read();    m=read();    for(int i=1;i<=m;i++){        a=read();        b=read();        x=read();        add(a,b,x);        add(b,a,x);    }    spfa(1,0);    spfa(n,1);    F(i,1,n) for(int j=head[i];j;j=e[j].f){        tmp=dis[i][0]+dis[e[j].t][1]+e[j].l;        if(tmp>dis[n][0]&&tmp<ans) ans=tmp;    }       printf("%d\n",ans);}

两次spfa分别从头和尾
之后枚举每一条边大于最小值的最小值

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二分答案
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