제출 #637686

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
637686		chjiao	Robot (JOI21_ho_t4)	C++17	100 / 100	685 ms	79032 KiB

Main

#include <bits/stdc++.h>
using namespace std;
#define ll  long long
#define newl '\n'
const ll mod = 1e9+7;
const ll MAX = 0x3f3f3f3f3f3f3f3f;
const ll MN = 1e5+1;
vector<map<int, ll>> colorsum(MN);
//vector<vector<pair<int,pair<int,ll>>>> neighbors(MN);
vector<map<int,vector<pair<int,ll>>>> neighbors(MN);
vector<ll> dp(MN, LLONG_MAX);
map<int, ll> dp2[MN];
struct quadriple{
    ll value;      //当前的value （cost）
    int curr;      // node
    // bool changed;  //是否改变
    int roadcolor; //来的road的color
    // ll roadvalue;  //来的road的value  （cost）
	int preCurr;   //来的路径
};

struct cmp {
    bool operator ()(quadriple & a, quadriple& b){
        return a.value > b.value;
    }
};

void dijkstra(ll s) {
    priority_queue<quadriple, vector<quadriple>, cmp> pq;
    pq.push({0LL,1,0, 0});
    dp[1] =0;
    while (!pq.empty()) {
        quadriple c = pq.top(); 
        pq.pop();
        
        if(c.roadcolor !=0 ){ //来的时候改变了颜色
            if(dp2[c.curr][c.roadcolor] != c.value){
                continue;
            }
            for(auto& [nPos, nValue]: neighbors[c.curr][c.roadcolor]){ //first = pos, second.first = color, second.second = num|| (was 'a')
                int nColor = c.roadcolor;
                // tie(nColor, nValue) = x;

               // if (c.roadcolor != nColor) continue;
                ll currsum = c.value +colorsum[c.curr][nColor] - nValue;
                if( currsum <  dp[nPos] ) { //只考虑改变其他颜色
                    dp[nPos] = currsum;
                    pq.push({currsum, nPos, 0, c.curr}) ;
                }
            }

        }else{  //来的时候未改变了颜色
            if(dp[c.curr] != c.value){
                continue;
            }
            for(auto& [nColor, x]: neighbors[c.curr]){ //first = pos, second.first = color, second.second = num|| (was 'a')
                for (auto & [nPos, nValue]: x) {
               // int nPos; ;ll nValue;
                

                    //禁止原路返回
                    if(nPos == c.preCurr) continue;

                    // //初始化DP2颜色
                    // if(dp2[nPos].count(nColor) == 0) dp2[nPos][nColor] = LLONG_MAX;

                    // case 1: 不改变nColor的颜色，但是改变其他所有相同颜色路径的颜色（含不改变任何颜色，但没有其他相同颜色路径时）
                    ll currsum = c.value + colorsum[c.curr][nColor] - nValue;
                    if(currsum < dp[nPos]){
                        dp[nPos] = currsum;
                        pq.push({currsum, nPos, 0, c.curr});
                    }

                    // case 2:  改变nColor的颜色，但是nColor的颜色和下一步的目的地颜色不同；
                    currsum = c.value + nValue;
                    if(currsum < dp[nPos]){
                        dp[nPos] = currsum;
                        pq.push({currsum, nPos, 0, c.curr});
                    }

                    // case 3: 改变了目的路径颜色，下一步目的地颜色相同并同步改变
                    if( dp2[nPos].count(nColor)==0 ||c.value < dp2[nPos][nColor] ){
                        pq.push({c.value, nPos, nColor, c.curr});
                        dp2[nPos][nColor] = c.value ;
                    }
                }
            }
        }
        
    }
}

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(0);
	// freopen("Robot.in", "r", stdin);
    // freopen("Robot.out", "w", stdout); 
    int N,M,a,b,c; 
    ll d;
    cin >> N >> M;
    for(int i=0; i<M; i++){
        cin >> a >> b >> c >> d;
        neighbors[a][c].push_back(make_pair(b,d));
        neighbors[b][c].push_back(make_pair(a,d));
        colorsum[a][c] +=d;  //presum
        colorsum[b][c]+=d;
    }
    dijkstra(0);
	
    cout << ((dp[N] <= 1e18)?dp[N] :-1 )<< endl;
}

• Subtask 1: 34 / 34
• Subtask 2: 24 / 24
• Subtask 3: 42 / 42