NOIP2013 货车运输 [LCA] [RMQ] [最小生成树] [ST倍增]

题目描述 Description
A 国有 n 座城市，编号从 1 到 n，城市之间有 m 条双向道路。每一条道路对车辆都有重量限制，简称限重。现在有 q 辆货车在运输货物，司机们想知道每辆车在不超过车辆限重的情况下，最多能运多重的货物。

输入描述 Input Description
第一行有两个用一个空格隔开的整数 n，m，表示 A 国有 n 座城市和 m 条道路。
接下来 m 行每行 3 个整数 x、y、z，每两个整数之间用一个空格隔开，表示从 x 号城市到 y 号城市有一条限重为 z 的道路。注意：x 不等于 y，两座城市之间可能有多条道路。
接下来一行有一个整数 q，表示有 q 辆货车需要运货。

接下来 q 行，每行两个整数 x、y，之间用一个空格隔开，表示一辆货车需要从 x 城市运输货物到 y 城市，注意：x 不等于 y。

输出描述 Output Description
输出共有 q 行，每行一个整数，表示对于每一辆货车，它的最大载重是多少。如果货车不能到达目的地，输出-1。

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首先看到这题， 由于要最大， 肯定是求最大生成树

那么 o(n2) dfs 求任意点对之间的最小边是可以想到的

但是看看数据范围肯定TLE

于是暴力链剖， 不过要注意query的时候判断的时候要 m+-1 但是递归下去要用m ， 可以画图举特例分析

1AC 代码：

#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<vector>#include<queue>#include<stack>#include<map>#include<set>#include<string>#include<iomanip>#include<ctime>#include<cctype>#include<algorithm>#ifdef WIN32#define AUTO "%I64d"#else#define AUTO "%lld"#endifusing namespace std;#define smin(x,tmp) x=min((x),(tmp))const int INF=0x3f3f3f3f;const int maxn=10005;const int maxm=50005;int fa[maxn];int find(int x) { return fa[x]==x?x:fa[x]=find(fa[x]); }inline bool union_find(int x,int y){    int t1=find(x),t2=find(y);    if(t1==t2) return false;    fa[t2]=t1;    return true;}map <pair<int,int>,int> g;struct Edge{    int to,next;    int val;}edge[maxm<<1];int head[maxn];int maxedge;inline void addedge(int u,int v,int c){    edge[++maxedge]=(Edge){v,head[u],c};    head[u]=maxedge;    edge[++maxedge]=(Edge){u,head[v],c};    head[v]=maxedge;    g[make_pair(u,v)]=c;    g[make_pair(v,u)]=c;}struct Road{    int from,to;    int cost;    bool operator < (const Road t) const    {        return cost>t.cost;// querying the Biggest MST !!!!    }}road[maxm];int n,m;int f[maxn],son[maxn],size[maxn],depth[maxn];int top[maxn],id[maxn],rid[maxn];int maxnode;//for segment tree (one-demensional array)int dfs1(int u,int father,int deep){    f[u]=father,size[u]=1,depth[u]=deep;    for(int i=head[u];~i;i=edge[i].next)    {        int v=edge[i].to;        if(v==father) continue;        size[u]+=dfs1(v,u,deep+1);        if(!son[u]||size[son[u]]<size[v]) son[u]=v;    }    return size[u];}void dfs2(int u,int tp){    top[u]=tp;id[u]=++maxnode;rid[maxnode]=u;// non de mixer la id-rid !!    if(son[u]) dfs2(son[u],tp);    for(int i=head[u];~i;i=edge[i].next)    {        int v=edge[i].to;        if(v==f[u]||v==son[u]) continue;        dfs2(v,v);    }}int tree[maxn<<2];//minvoid build(int root,int l,int r){    if(r-l==1)    {        tree[root]=g[make_pair(rid[l],rid[r])];        return;    }    int m=(l+r)>>1;    build(root<<1,l,m);    build(root<<1|1,m,r);    tree[root]=min(tree[root<<1],tree[root<<1|1]);}int query(int root,int l,int r,int x,int y)//query min{    if(x==y) return INF;//must!! when query the same node !!!!!!    if(x<=l&&r<=y) return tree[root];    int m=(l+r)>>1;    int t1=INF,t2=INF;    if(x<=m-1&&l<=y) t1=query(root<<1,l,m,x,y);//here too, use x<=m-1 in case stucking at m    if(y>=m+1&&r>=x) t2=query(root<<1|1,m,r,x,y);//be conscious of m or m+1, query m but judge m+1    return min(t1,t2);}int Find(int u,int v)//find min{    int t1=top[u],t2=top[v];    int ret=INF;    while(t1^t2)    {        if(depth[t1]<depth[t2]) swap(t1,t2),swap(u,v);        smin(ret,query(1,1,maxnode,id[t1],id[u]));        smin(ret,g[make_pair(t1,f[t1])]);        u=f[t1];        t1=top[u];    }    if(depth[u]<depth[v]) swap(u,v);    return min(ret,query(1,1,maxnode,id[v],id[u]));}inline void init(){    scanf("%d%d",&n,&m);    for(int i=1;i<=m;i++) scanf("%d%d%d",&road[i].from,&road[i].to,&road[i].cost);    for(int i=1;i<=n;i++) fa[i]=i;    memset(head,-1,sizeof(head));    maxedge=-1;}void kruskal(){    sort(road+1,road+m+1);    int pos=1,tot=0;    while(pos<=m && tot^(n-1))    {        if(union_find(road[pos].from,road[pos].to)) tot++,addedge(road[pos].from,road[pos].to,road[pos].cost);        pos++;    }}bool vis[maxn];//for union_findvoid build_forest(){    for(int i=1;i<=n;i++)    {        int father=find(i);        if(!vis[father])        {            vis[father]=true;            dfs1(father,0,1);            dfs2(father,father);        }    }    build(1,1,maxnode);}int main(){    freopen("truck.in","r",stdin);    freopen("truck.out","w",stdout);    init();    kruskal();    build_forest();    int q;    scanf("%d",&q);    while(q--)    {        int x,y;        scanf("%d%d",&x,&y);        if(find(x)^find(y)) printf("-1\n");        else printf("%d\n",Find(x,y));    }    return 0;}

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但是NOIP正解一定不是链剖， 此题要用到 LCA 的 ST 倍增算法， 并且属于精确的查询，没有重叠部分， 可以用sum等进行替换， 边dfs边更新

用 f[i][j] = f[f[i][j-1]][j-1] 保存 i 前面的第2^i 节点， 与普通 RMQ 不同

用 dp[i][j] = min ( dp[i][j-1] , dp[f[i][j-1]][j-1] ) 来维护倍增的最小值 （ 当然 sum 也一样 ， 因为精确求范围， 满足区间加 ）

AC代码：

#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<vector>#include<queue>#include<stack>#include<map>#include<set>#include<string>#include<iomanip>#include<ctime>#include<cctype>#include<algorithm>#ifdef WIN32#define AUTO "%I64d"#else#define AUTO "%lld"#endifusing namespace std;#define smin(x,tmp) x=min((x),(tmp))#define smax(x,tmp) x=max((x),(tmp))const int INF=0x3f3f3f3f;const int maxn=10005;const int maxm=50005;const int maxd=20;int fa[maxn];int find(int x) { return fa[x]==x?x:fa[x]=find(fa[x]); }inline bool union_find(int x,int y){    int t1=find(x),t2=find(y);    if(t1==t2) return false;    fa[t2]=t1;    return true;}struct Edge{    int to,next;    int val;}edge[maxm<<1];int head[maxn];int maxedge;inline void addedge(int u,int v,int c){    edge[++maxedge]=(Edge){v,head[u],c};    head[u]=maxedge;    edge[++maxedge]=(Edge){u,head[v],c};    head[v]=maxedge;}struct Road{    int from,to;    int cost;    bool operator < (const Road t) const    {        return cost>t.cost;    }}road[maxm];int n,m;inline void init(){    scanf("%d%d",&n,&m);    for(int i=1;i<=m;i++) scanf("%d%d%d",&road[i].from,&road[i].to,&road[i].cost);    for(int i=1;i<=n;i++) fa[i]=i;    memset(head,-1,sizeof(head));    maxedge=-1;}void kruskal(){    sort(road+1,road+m+1);    int pos=1,tot=0;    while(pos<=m && tot^(n-1))    {        if(union_find(road[pos].from,road[pos].to)) tot++,addedge(road[pos].from,road[pos].to,road[pos].cost);        pos++;    }}int f[maxn][maxd+5],dp[maxn][maxd+5]; // the node 2^j after u int depth[maxn];void dfs(int u,int deep){    depth[u]=deep;    for(int k=1;(1<<k)<=n;k++)    {        f[u][k] = f[f[u][k-1]][k-1];        dp[u][k] = min(dp[u][k-1] , dp[f[u][k-1]][k-1]);    }    for(int i=head[u];~i;i=edge[i].next)    {        int v=edge[i].to;        if(!depth[v])        {            f[v][0]=u;dp[v][0]=edge[i].val;//initialize here rather than in the former context!! no considering the roor coz INF is INF, not changing and not visiting the value!!            dfs(v,deep+1);        }    }}int Find(int u,int v){    int ans=INF;    if(depth[u] < depth[v]) swap(u,v);// making the u is deeper!!    // make u v at the same depth    for(int k=maxd;k>=0;k--) // k>=0 here!! or cannot jump at the same depth!!!        if(depth[v] <= depth[f[u][k]]) // f[0] = 0, indicates its beyond the root!!        {            smin( ans , dp[u][k] );            u=f[u][k];        }    if(u == v) return ans; //special judge of one of them is the LCA    // jump at the same time     for(int k=maxd;k>=0;k--) // k>=0 here too!!        if(f[u][k] ^ f[v][k])        {            smin(ans , min( dp[u][k] , dp[v][k] ) );            u = f[u][k];            v = f[v][k];        }    return min( ans , min(dp[u][0] , dp[v][0])); // u v this time is the F1 of LCA}int main(){    freopen("truck.in","r",stdin);    freopen("truck.out","w",stdout);    init();    kruskal();    memset(dp,0x3f,sizeof(dp));    for(int i=1;i<=n;i++)        if(!depth[i]) dfs(i,1);    int q;    scanf("%d",&q);    while(q--)    {        int x,y;        scanf("%d%d",&x,&y);        if(find(x)^find(y)) printf("-1\n");        else printf("%d\n",Find(x,y));    }    return 0;}

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顺便写了一个利用dfn求LCA的方法 HDU 2874

#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<vector>#include<queue>#include<stack>#include<map>#include<set>#include<string>#include<iomanip>#include<ctime>#include<climits>#include<cctype>#include<algorithm>#ifdef WIN32#define AUTO "%I64d"#else#define AUTO "%lld"#endifusing namespace std;const int maxn=10005;struct Edge{    int to,next;    int val;}edge[maxn*maxn];int head[maxn];int maxedge;inline void addedge(int u,int v,int c){    edge[++maxedge]=(Edge){v,head[u],c};    head[u]=maxedge;    edge[++maxedge]=(Edge){u,head[v],c};    head[v]=maxedge;}int n,m,q;int fa[maxn];int find(int x) { return fa[x]==x?x:fa[x]=find(fa[x]); }bool union_find(int x,int y){    int t1=find(x),t2=find(y);    if(t1==t2) return false;    fa[t2]=t1;    return true;}int maxnode;int dfn[maxn],ver[maxn<<1];//dfn: first visit maxnode, ver: reverse function of dfn, indicating the number of vertexint depth[maxn<<1],dis[maxn];//depth: the depth of dfn, dis: from root to vertexinline bool init(){    if(!~scanf("%d%d%d",&n,&m,&q)) return false;    for(int i=1;i<=n;i++) fa[i]=i;    memset(head,-1,sizeof(head));    memset(dfn,0,sizeof(dfn));    maxedge=-1;maxnode=0;    for(int i=1;i<=m;i++)    {        int u,v,c;        scanf("%d%d%d",&u,&v,&c);        addedge(u,v,c);        union_find(u,v);    }    return true;}void dfs(int u,int deep){    dfn[u]=++maxnode;ver[maxnode]=u;depth[maxnode]=deep;    for(int i=head[u];~i;i=edge[i].next)    {        int v=edge[i].to;        if(dfn[v]) continue;        dis[v]=dis[u]+edge[i].val;        dfs(v,deep+1);        depth[++maxnode]=deep;ver[maxnode]=u;    }}const int maxdepth=20;int dp[maxn<<1][maxdepth];void ST(int n)// n=::maxnode{    for(int i=1;i<=n;i++) dp[i][0]=i;    for(int j=1;(1<<j)<=n;j++)//careful of the limits        for(int i=1;i+(1<<j)-1<=n;i++)//careful of the limits by y=dp[i+(1<<j-1)][j-1] and the limit of i+(1<<j)-1<=n, must -1 'coz the i+(1<<j)-1 is possible to be n!!        {            int x=dp[i][j-1],y=dp[i+(1<<j-1)][j-1];            dp[i][j]=depth[x]<depth[y]?x:y;        }}inline int RMQ(int l,int r){    int k=0;    while(1<<(k+1)<=r-l+1) k++;// careful of the limits!!    int x=dp[l][k],y=dp[r-(1<<k)+1][k];    return depth[x]<depth[y]?x:y;}inline int LCA(int u,int v){    int x=dfn[u],y=dfn[v];    if(x>y) swap(x,y);    int root=RMQ(x,y);    return ver[root];}int main(){    freopen("city.in","r",stdin);    freopen("city.out","w",stdout);    while(init())    {        for(int i=1;i<=n;i++)            if(!dfn[i]) dfs(i,1);        ST(maxnode);        while(q--)        {            int x,y;            scanf("%d%d",&x,&y);            if(find(x)^find(y)) printf("Not connected\n");            else printf("%d\n",dis[x]+dis[y]-(dis[LCA(x,y)]<<1));        }    }    return 0;}