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  • 1204C

    1204C - Anna, Svyatoslav and Maps

    看别人题解才知道题目是怎么回事。。。

    #include <iostream>
    #include <vector>
    #include <algorithm>
    #include <string>
    #include <set>
    #include <queue>
    #include <map>
    #include <sstream>
    #include <cstdio>
    #include <cstring>
    #include <numeric>
    #include <cmath>
    #include <iomanip>
    #include <deque>
    #include <bitset>
    //#include <unordered_set>
    //#include <unordered_map>
    #define ll              long long
    #define pii             pair<int, int>
    #define rep(i,a,b)      for(ll  i=a;i<=b;i++)
    #define dec(i,a,b)      for(ll  i=a;i>=b;i--)
    #define forn(i, n)      for(ll i = 0; i < int(n); i++)
    using namespace std;
    int dir[4][2] = { { 1,0 },{ 0,1 } ,{ 0,-1 },{ -1,0 } };
    const long long INF = 0x7f7f7f7f7f7f7f7f;
    const int inf = 0x3f3f3f3f;
    const double pi = 3.14159265358979323846;
    const double eps = 1e-6;
    const int mod = 998244353;
    const int N = 1e6 + 5;
    //if(x<0 || x>=r || y<0 || y>=c)
    
    inline ll read()
    {
        ll x = 0; bool f = true; char c = getchar();
        while (c < '0' || c > '9') { if (c == '-') f = false; c = getchar(); }
        while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
        return f ? x : -x;
    }
    ll gcd(ll m, ll n)
    {
        return n == 0 ? m : gcd(n, m % n);
    }
    ll lcm(ll m, ll n)
    {
        return m * n / gcd(m, n);
    }
    bool prime(int x) {
        if (x < 2) return false;
        for (int i = 2; i * i <= x; ++i) {
            if (x % i == 0) return false;
        }
        return true;
    }
    inline int qpow(int x, ll n) {
        int r = 1;
        while (n > 0) {
            if (n & 1) r = 1ll * r * x % mod;
            n >>= 1; x = 1ll * x * x % mod;
        }
        return r;
    }
    inline int add(int x, int y) {
        return ((x%mod)+(y%mod))%mod;
    }
    inline int sub(int x, int y) {
        x -= y;
        return x < 0 ? x += mod : x;
    }
    inline int mul(int x, int y) {
        return (1ll * (x %mod) * (y % mod))%mod;
    }
    inline int Inv(int x) {
        return qpow(x, mod - 2);
    }
    
    int main()
    {
        int n;
        cin >> n;
        vector<vector<int>> g(n+1,vector<int>(n+1,inf));
        rep(i, 1, n)
        {
            string s;
            cin >> s;
            forn(j, n)
            {
                if (s[j] == '1')
                {
                    g[i][j + 1] = 1;
                }
            }
        }
        rep(i, 0, n)
            g[i][i] = 0;
        int len;
        cin >> len;
        vector<int> a(len + 1);
        rep(i, 1, len)
            cin >> a[i];
        rep(i, 1, n)
        {
            rep(j, 1, n)
            {
                rep(k, 1, n)
                {
                    g[j][k] = min(g[j][i] + g[i][k], g[j][k]);
                }
            }
        }
        vector<int> res;
        res.push_back(a[1]);
        int dis = 0;
        rep(i, 2, len)
        {
            dis += g[a[i - 1]][a[i]];
            if (dis > g[res.back()][a[i]])
            {
                res.push_back(a[i - 1]);
                dis = g[res.back()][a[i]];
            }
        }
        res.push_back(a[len]);
        cout << res.size() << endl;
        rep(i, 0, res.size()-1)
            cout << res[i] << " ";
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/dealer/p/13282024.html
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