CF1296F Berland Beauty (Tree, dfs/bfs)

题目链接

题解:

走一遍dfs/bfs得到u->v的路径。每一个m,保证走过得边>=此最小值。再走一遍m,看路径的最小值是否是提供的。

#include <bits/stdc++.h>
# define LL long long
using namespace std;

int n;
int m;
void bfs(int u, vector<pair<int,int>> &p, vector<vector<pair<int,int>>> &adj){
    queue<int> q;
    q.push(u);
    while(!q.empty()){
        int cur=q.front();
        q.pop();
        for(auto as:adj[cur]){
            int v=as.first;
            if(v==p[cur].first) continue;
            int e=as.second;

            p[v].first=cur;
            p[v].second=e;
            q.push(v);
        }
    }
}

int main(){
    scanf("%d", &n);
    vector<vector<pair<int,int>>> adj(n+1);
    for(int i=1;i<=n-1;++i){
        int u,v;
        scanf("%d %d", &u, &v);
        adj[u].push_back(make_pair(v,i));
        adj[v].push_back(make_pair(u,i));
    }

    vector<vector<pair<int,int>>> pre(n+1,vector<pair<int,int>>(n+1));
    for(int i=1;i<=n;++i){
        pre[i][i]=make_pair(0,0);
        bfs(i,pre[i], adj);
    }

    scanf("%d", &m);
    vector<int> val(n,0);
    vector<vector<int>> query(m+1,vector<int>(3,0));
    for(int i=1;i<=m;++i){
        scanf("%d %d %d", &query[i][0], &query[i][1], &query[i][2]);
    }
    for(int i=1;i<=m;++i){
        int u=query[i][0];
        int v=query[i][1];
        int f=query[i][2];
        while(true){
            int pree=pre[u][v].second;
            int prev=pre[u][v].first;
            val[pree]=max(val[pree],f);
            v=prev;
            if(v==u) break;
        }
    }

    for(int i=1;i<=m;++i){
        int u=query[i][0];
        int v=query[i][1];
        int f=query[i][2];
        int mx=1000000+10;
        while(true){
            int pree=pre[u][v].second;
            int prev=pre[u][v].first;
            mx=min(mx,val[pree]);
            v=prev;
            if(v==u) break;
        }
        if(mx!=f){
            printf("-1");
            return 0;
        }
    }
    for(int i=1;i<=n-1;++i){
        if(val[i]==0) printf("100 ");
        else printf("%d ", val[i]);
    }
    return 0;
}

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转载自www.cnblogs.com/FEIIEF/p/12263876.html