HDU 5385 The path

如果我们知道每个点的dis值和最短路径树的话，方案是很容易构造的

我们可以采取贪心做法，一开始将1号点作为最短路径树的根，然后左边从2开始，右边从n开始，只要之前加入的点有边连向他们就加入

这样一个点加入的时间就是他的dis值，最短路径树上的父亲也可以确定，于是输出时非树边长度为n，树边长度为两个端点dis之差。

//      whn6325689//      Mr.Phoebe//      http://blog.csdn.net/u013007900#include <algorithm>#include <iostream>#include <iomanip>#include <cstring>#include <climits>#include <complex>#include <fstream>#include <cassert>#include <cstdio>#include <bitset>#include <vector>#include <deque>#include <queue>#include <stack>#include <ctime>#include <set>#include <map>#include <cmath>#include <functional>#include <numeric>#pragma comment(linker, "/STACK:1024000000,1024000000")using namespace std;#define eps 1e-9#define PI acos(-1.0)#define INF 0x3f3f3f3f#define LLINF 1LL<<62#define speed std::ios::sync_with_stdio(false);typedef long long ll;typedef unsigned long long ull;typedef long double ld;typedef pair<ll, ll> pll;typedef complex<ld> point;typedef pair<int, int> pii;typedef pair<pii, int> piii;typedef vector<int> vi;#define CLR(x,y) memset(x,y,sizeof(x))#define CPY(x,y) memcpy(x,y,sizeof(x))#define clr(a,x,size) memset(a,x,sizeof(a[0])*(size))#define cpy(a,x,size) memcpy(a,x,sizeof(a[0])*(size))#define mp(x,y) make_pair(x,y)#define pb(x) push_back(x)#define lowbit(x) (x&(-x))#define MID(x,y) (x+((y-x)>>1))template<class T>inline bool read(T &n){    T x = 0, tmp = 1;    char c = getchar();    while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();    if(c == EOF) return false;    if(c == '-') c = getchar(), tmp = -1;    while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar();    n = x*tmp;    return true;}template <class T>inline void write(T n){    if(n < 0)    {        putchar('-');        n = -n;    }    int len = 0,data[20];    while(n)    {        data[len++] = n%10;        n /= 10;    }    if(!len) data[len++] = 0;    while(len--) putchar(data[len]+48);}//-----------------------------------const int MAXN=100010;set<int> st;int n,m;int vis[MAXN],cost[MAXN];struct Edge{    int to,next;}e[MAXN<<1];int head[MAXN],tot;int fa[MAXN],dis[MAXN],idx[MAXN];void init(){    CLR(fa,-1);CLR(head,-1);    CLR(vis,0);CLR(cost,0);    tot=0;st.clear();}void addedge(int u,int v){    e[tot].next=head[u];    e[tot].to=v;    head[u]=tot++;}int main(){    int T;    read(T);    while(T--)    {        init();        read(n),read(m);        for(int i=0,u,v;i<m;i++)        {            read(u),read(v);              addedge(u,v);        }        int cnt=1;        int now=1,l=1,r=n,v;        fa[1]=1;dis[1]=0;        while(cnt<n)        {            for(int i=head[now];~i;i=e[i].next)            {                v=e[i].to;                if(fa[v]!=-1 || st.find(v)!=st.end()) continue;                fa[v]=now;idx[v]=i;                st.insert(v);            }            //cout<<" :"<<now<<endl;            cnt++;            int be=*st.begin();            if(be==l+1)            {                cost[idx[be]]=cnt-dis[fa[be]]-1;                dis[be]=cnt-1;                vis[idx[be]]=true;                st.erase(be);                l=now=be;                //cout<<"1:"<<be<<endl;            }            else            {                be=*(st.rbegin());                cost[idx[be]]=cnt-dis[fa[be]]-1;                dis[be]=cnt-1;                vis[idx[be]]=true;                st.erase(be);                now=be;                //cout<<"2:"<<be<<endl;            }        }        for(int i=0;i<m;i++)            if(!vis[i])                printf("%d\n",n);            else                printf("%d\n",cost[i]);        }    return 0;}

换一种姿势，总体思路差不多

#include<cstdio>#include<cstring>#include<cstdlib>#include<iostream>#include<vector>#include<set>#define maxn 100008using namespace std;struct yyy{    int x,m;};int ans[maxn];vector<yyy> g[maxn];set<int> s;int vis[maxn];int pre[maxn];int prej[maxn];int main(){    int T;    int n,m,i;    scanf("%d",&T);    while (T--)    {        scanf("%d%d",&n,&m);        for (i = 1 ; i <= n ; i++)        {            g[i].clear();            vis[i] = -1;            pre[i] = 0;            prej[i] = 0;        }        int x,y,cnt = 0;        for (i = 1 ; i <= m ; i++)        {            ans[i] = n;            scanf("%d%d",&x,&y);            g[x].push_back( (yyy) {y,++cnt} );        }        int l = 1,r = n;        int v;        s.clear();        s.insert(1);        vis[1] = 0;        int dis = -1;        while ( !s.empty() )        {            int L = *s.begin(),R = *(--s.end());           // cout<<L<<' '<<R<<endl;            if (L == l)                x = L,l++;            else x = R,r--;            s.erase(x);            dis++;            if (x != 1)            {                ans[pre[x]] = dis - vis[prej[x]];                vis[x] = dis;            }            for (i = 0 ; i < g[x].size() ; i++)                {                    v = g[x][i].x;                    if ( vis[v] == -1 )                    {                        vis[v] = 0;                        pre[v] = g[x][i].m;                        prej[v] = x;                        s.insert(v);                    }                }        }        for (i = 1 ; i <= m ; i++)            printf("%d ",ans[i]);        printf("\n");    }    return 0;}