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  • 比赛-Round 2 (10 Jul)

    谢谢 hzwer 学长!

    1. 小奇挖矿

    从右往左决策消除后效性。

     1 #include <cstdio>
     2 #include <algorithm>
     3 
     4 using namespace std;
     5 
     6 const int _N = 120000;
     7 
     8 int type[_N], A[_N];
     9 
    10 int main()
    11 {
    12     double ans = 0, P, k1, k2, K, C;
    13     int N, i;
    14     scanf("%d%lf%lf%lf", &N, &K, &C, &P);
    15     k1 = 1.0-K/100.0;
    16     k2 = 1.0+C/100.0;
    17     for (i = 1; i <= N; ++i)
    18         scanf("%d%d", &type[i], &A[i]);
    19     for (i = N; i >= 1; --i) {
    20         if (type[i] == 1)
    21             ans = max(ans, ans*k1+A[i]);
    22         else
    23             ans = max(ans, ans*k2-A[i]);
    24     }
    25     printf("%.2lf
    ", ans*P);
    26     return 0;
    27 }
    View Code

    2. 小奇的数列

    3. 小奇回地球

    二分答案,判断到终点路上时候有负环。

    #include <stdio.h>
    #include <queue>
    #include <vector>
    #include <algorithm>
    #include <string.h>
    
    using namespace std;
    
    const int _N = 120;
    const int INF = 1e9;
    
    struct edge {
        int v, w;
        edge(int v, int w):
            v(v), w(w) { }
    };
    
    queue<int> Q;
    vector<edge> G[_N];
    int N, M;
    int dis[_N], mk[_N], cnt[_N];
    bool GG[_N][_N];
    
    void Ins(int x, int y, int w) { G[x].push_back(edge(y, w)); return; }
    
    bool Check(int ex)
    {
        int i;
        for (i = 1; i <= N; ++i)
            dis[i] = INF, mk[i] = false, cnt[i] = 0, GG[i][i] = true;
        while (!Q.empty()) Q.pop();
        Q.push(1), dis[1] = 0, cnt[1] = 1, mk[1] = true;
        while (!Q.empty()) {
            int p = Q.front();
            Q.pop(), mk[p] = false;
            for (i = G[p].size()-1; i >= 0; --i) {
                edge v = G[p][i];
                if (!GG[v.v][N] || !GG[1][v.v]) continue;
                if (dis[v.v] > dis[p] + v.w + ex) {
                    dis[v.v] = dis[p] + v.w + ex;
                    if (!mk[v.v]) {
                        Q.push(v.v);
                        if (++cnt[v.v] > N) return false;
                    }
                }
            }
        }
        return dis[N] >= 0 && dis[N] < INF;
    }
    
    int main()
    {
        int T;
        scanf("%d", &T);
        while (T--) {
            //CLEAR
            int bi_r, bi_l, i, j, k, ans = -1;
            scanf("%d%d", &N, &M);
            
            for (i = 1; i <= N; ++i) G[i].clear();
            memset(GG, 0, sizeof GG);
            
            for (i = 1; i <= M; ++i) {
                int x, y, w;
                scanf("%d%d%d", &x, &y, &w);
                GG[x][y] = true;
                Ins(x, y, w);
            }
            
            for (k = 1; k <= N; ++k)
                for (i = 1; i <= N; ++i)
                    for (j = 1; j <= N; ++j)
                        if (GG[i][k] && GG[k][j]) GG[i][j] = true;
                                
            bi_l = -120000, bi_r = 120000;
            while (bi_l <= bi_r) {
                int bi_mid = bi_l+bi_r >> 1;
                if (Check(bi_mid))
                    ans = dis[N], bi_r = bi_mid-1;
                else
                    bi_l = bi_mid+1;
            }
            
            printf("%d
    ", ans);
        }
        return 0;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/ghcred/p/9293620.html
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