[Codeforces 507E] Breaking Good

[题目链接]

         https://codeforces.com/contest/507/problem/E

[算法]

        首先BFS求出1到其余点的最短路 ， N到其余点的最短路，记为distA[]和distB[]

        显然 ， 我们只需最大化求出的最短路上没有被破坏的边即可 ， 不妨用f[i]表示现在在城市i , distA[i] + distB[i] = distA[N] ， 最多还能经过几条没有被破坏的边 

        记忆化搜索即可

        时间复杂度 ： O(N)

[代码]

        

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 10;
const int inf = 1e9;

struct edge
{
        int to , state , id , nxt;
} e[MAXN << 1];
struct info
{
        int u , v , ns;
} res[MAXN];

int n , m , tot;
int u[MAXN],v[MAXN],state[MAXN],head[MAXN],dista[MAXN],distb[MAXN],f[MAXN];
pair<int,int> nxt[MAXN];
bool visited[MAXN];

template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); }
template <typename T> inline void chkmin(T &x,T y) { x = min(x,y); }
template <typename T> inline void read(T &x)
{
    T f = 1; x = 0;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
    for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
    x *= f;
}
inline void addedge(int u,int v,int s,int id)
{
        tot++;
        e[tot] = (edge){v,s,id,head[u]};
        head[u] = tot; 
} 
inline void bfs1(int s)
{
        queue< int > q;
        dista[s] = 0;
        visited[s] = true;
        q.push(s);
        while (!q.empty())        
        {
                int cur = q.front();
                q.pop();
                for (int i = head[cur]; i; i = e[i].nxt)
                {
                        int v = e[i].to;
                        if (!visited[v])
                        {
                                visited[v] = true;
                                dista[v] = dista[cur] + 1;
                                q.push(v);
                        }
                }
        }
}
inline void bfs2(int s)
{
        queue< int > q;
        distb[s] = 0;
        visited[s] = true;
        q.push(s);
        while (!q.empty())
        {
                int cur = q.front();
                q.pop();
                for (int i = head[cur]; i; i = e[i].nxt)
                {
                        int v = e[i].to;
                        if (!visited[v])
                        {
                                visited[v] = true;
                                distb[v] = distb[cur] + 1;
                                q.push(v);
                        }
                }
        }
}
inline int dp(int u)
{
        if (u == n) return f[u] = 0;
        if (f[u] != -1) return f[u];
        f[u] = -inf;
        for (int i = head[u]; i; i = e[i].nxt)
        {
                int v = e[i].to , st = e[i].state , id = e[i].id , value;
                if (dista[u] + distb[v] + 1 != dista[n]) continue;
                if (st == 1) 
                {
                        value = dp(v) + 1;
                        if (value > f[u]) 
                        {
                                f[u] = value;
                                nxt[u] = make_pair(v,id);
                        }
                } else 
                {
                        value = dp(v);
                        if (value > f[u])
                        {
                                f[u] = value;
                                nxt[u] = make_pair(v,id);
                        }
                }
        }
        return f[u];
}
inline void getpath(vector<int> &a)
{
        int now = 1;
        while (nxt[now].first)
        {
                a.push_back(nxt[now].second);
                now = nxt[now].first;        
        }        
}

int main()
{
        
        read(n); read(m);
        for (int i = 1; i <= m; i++)
        {
                read(u[i]); read(v[i]); read(state[i]);
                addedge(u[i],v[i],state[i],i);    
                addedge(v[i],u[i],state[i],i);    
        }
        memset(visited,false,sizeof(visited));
        bfs1(1);
        memset(visited,false,sizeof(visited));
        bfs2(n);
        memset(f,255,sizeof(f));
        dp(1);
        vector<int> path;
        getpath(path);
        int len = 0;
        memset(visited,false,sizeof(visited));
        for (unsigned i = 0; i < path.size(); i++)
        {
                visited[path[i]] = true;
                if (state[path[i]] == 0) res[++len] = (info){u[path[i]],v[path[i]],1};        
        }
        for (int i = 1; i <= m; i++)
        {
                if (!visited[i] && state[i])
                        res[++len] = (info){u[i],v[i],0};
        }
        printf("%d\n",len);
        for (int i = 1; i <= len; i++) printf("%d %d %d\n",res[i].u,res[i].v,res[i].ns);
        
        return 0;
    
}