CodeForces 507E Breaking Good 2维权重dij

Breaking Good

题解：

        2维权重dij， 先距离最短， 后改变最小。

 

在这个题中， 如果要改变最小， 则让更多的可用边放进来。

然后可以用pre存下关键边。

 

代码：

#include<bits/stdc++.h>
using namespace std;
#define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
#define LL long long
#define ULL unsigned LL
#define fi first
#define se second
#define pb push_back
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lch(x) tr[x].son[0]
#define rch(x) tr[x].son[1]
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
typedef pair<int,int> pll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL _INF = 0xc0c0c0c0c0c0c0c0;
const LL mod =  (int)1e9+7;
const int N = 1e5 + 100;
vector<pll> vc[N];
struct Node{
    int o, u, d, p;
    bool operator < (const Node & x) const{
        if(d == x.d) return p > x.p;
        return d > x.d;
    }
};
int pre[N];
int vis[N];
priority_queue<Node> pq;
void dij(){
    pq.push({0, 1, 0, 0});
    while(!pq.empty()){
        Node t = pq.top();
        pq.pop();
        if(vis[t.u]) continue;
        vis[t.u] = 1;
        pre[t.u] = t.o;
        for(pll tmp : vc[t.u]){
            if(vis[tmp.fi]) continue;
            pq.push({t.u, tmp.fi, t.d+1, t.p + !tmp.se});
        }
    }
}
vector<Node> ans;
int fpre[N];
int main(){
    int n, m, u, v, op;
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= m; ++i){
        scanf("%d%d%d", &u, &v, &op);
        vc[u].pb(pll(v, op));
        vc[v].pb(pll(u, op));
    }
    dij();
    int x = n;
    while(x){
        fpre[x] = pre[x];
        x = pre[x];
    }
    for(int i = 1; i <= n; ++i){
        for(pll &t : vc[i]){
            if(t.fi < i){
                int v = t.fi;
                if((fpre[i] == v || fpre[v] == i)){
                    if(!t.se)
                        ans.pb({v, i, 1, 0});
                }
                else if(t.se)
                    ans.pb({v, i, 0, 0});
            }
        }
    }
    printf("%d
", ans.size());
    for(Node & t : ans){
        printf("%d %d %d
", t.o, t.u, t.d);
    }
    return 0;
}