高效算法设计(五)

本节内容再深入贪心算法和算法设计
其中会用到一些数学证明
辅助证明一些性质

然后在这些性质的基础上
进行贪心选择

构造法

LA4378

const int maxn = 100000 + 10;

class Data {
public:
    int val, id;
    Data(int v = 0, int i = 0) : val(v), id(i) {}
};

bool cmp(const Data& lhs, const Data& rhs) {
    return lhs.val > rhs.val;
}

Data data[maxn];
int ans[maxn];
int n;

void init() {
    Set(ans, 0);
}

int main() {
    freopen("input.txt", "r", stdin);
    while (~scanf("%d", &n)) {
        init();
        llong sum = 0;
        _for(i, 0, n) {
            scanf("%d", &data[i].val);
            data[i].id = i;
            sum += data[i].val;
        }

        if(sum % 2) {
            printf("No\n");
            continue;
        }
        else sum /= 2;

        sort(data, data + n, cmp);
        _for(i, 0, n) {
            if(data[i].val <= sum) {
                sum -= data[i].val;
                ans[data[i].id] = 1;
            }
            else ans[data[i].id] = -1;
        }

        printf("Yes\n");
        _for(i, 0, n - 1) printf("%d ", ans[i]);
        printf("%d\n", ans[n - 1]);
    }
}

环形排列

UVA10570

const int maxn = 500 + 10;
int A[maxn], B[maxn], pos[maxn];
int N;

void init() {
    Set(A, 0);
    Set(B, 0);
    Set(pos, 0);
}

void copy(int st, int step) {
    _for(i, 0, N) {
        B[i] = A[(st + step * i + N) % N];
        pos[B[i]] = i;
    }
}

int solve() {
    int ans = 0;
    _for(i, 0, N) {
        int p = pos[i];
        if(p == i) continue;
        pos[B[i]] = p;
        pos[i] = i;
        swap(B[i], B[p]);
        ans++;
    }
    return ans;
}

int main() {
    freopen("input.txt", "r", stdin);
    while (scanf("%d", &N) == 1 && N) {
        init();

        _for(i, 0, N) {
            scanf("%d", &A[i]);
            A[i]--;
        }

        int res = maxn;
        _for(i, 0, N) {
            copy(i, 1);
            res = min(res, solve());
            copy(i, -1);
            res = min(res, solve());
        }
        printf("%d\n", res);
    }
}

图上的贪心算法

LA4993

const int maxn = 10000 + 10;

class Edge {
public:
    int from, to;
    Edge(int f = 0, int t = 0) : from(f), to(t) {}
};

vector<Edge> edges;
vector<int> G[maxn];
int n, m, k;

int color[maxn], vis[maxn];

void init() {
    Set(color, 0);
    Set(vis, 0);
    _for(i, 0, maxn) G[i].clear();
    edges.clear();
    k = 0;
}

void initVis() {
    Set(vis, 0);
}

void addEdge(int from, int to) {
    edges.push_back(Edge(from, to));
    int m = edges.size();
    G[from].push_back(m - 1);
}

void dfs(int u) {
    initVis();
    _for(i, 0, G[u].size()) {
        Edge cur = edges[G[u][i]];
        if(color[cur.to]) vis[color[cur.to]] = 1;
    }

    _rep(i, 1, k) {
        if(!vis[i]) {
            color[u] = i;
            break;
        }
    }

    _for(i, 0, G[u].size()) {
        Edge cur = edges[G[u][i]];
        if(!color[cur.to]) dfs(cur.to);
    }
}

int main() {
    freopen("input.txt", "r", stdin);
    int kase = 0;

    while (scanf("%d%d", &n, &m) != EOF) {
        init();
        if(++kase > 1) printf("\n");

        int u, v;
        _for(i, 0, m) {
            scanf("%d%d", &u, &v);
            addEdge(u - 1, v - 1);
            addEdge(v - 1, u - 1);
        }
        _for(i, 0, n) k = max(k, (int)G[i].size());

        if(k % 2 == 0) k++;
        cout << k << endl;

        dfs(0);
        _for(i, 0, n) cout << color[i] << endl;
    }
}

枚举排列

处理小数点的方法

int readint() {
    char buf[8];
    scanf("%s", buf);

    int a = 0, b = 0;
    sscanf(buf, "%d", &a);

    char* pp = strchr(buf, '.');
    if(pp) sscanf(++pp, "%d", &b);

    return a * 100 + b;
}

example

LA3664

const int maxn = 16384 + 5;
vector<int> per[maxn];
vector<int> rk;
int n;

void init() {
    _for(i, 0, maxn) per[i].clear();
    rk.clear();
}

int readint() {
    char buf[8];
    scanf("%s", buf);

    int a = 0, b = 0;
    sscanf(buf, "%d", &a);

    char* pp = strchr(buf, '.');
    if(pp) sscanf(++pp, "%d", &b);

    return a * 100 + b;
}

int read() {
    int x;
    scanf("%d", &x);
    return x;
}

int solve() {
    int lastScore = -1, lastID = -1;
    _for(i, 0, rk.size()) {
        int id = rk[i];
        if(lastScore == -1) lastScore = per[id].front();
        else {
            bool found = false;
            _for(j, 0, per[id].size()) {
                if(per[id][j] < lastScore || (per[id][j] == lastScore && id > lastID)) {
                    lastScore = per[id][j];
                    found = true;
                    break;
                }
            }
            if(!found) return -1;
        }
        lastID = id;
    }
    return lastScore;
}

int main() {
    freopen("input.txt", "r", stdin);
    for(int kase = 1; scanf("%d", &n) == 1 && n; kase++) {
        init();

        _for(i, 0, n) {
            per[i].clear();

            per[i].push_back(0);
            _for(j, 0, 3) per[i].push_back(readint());

            per[i].push_back(per[i][1] + per[i][2]);
            per[i].push_back(per[i][1] + per[i][3]);
            per[i].push_back(per[i][2] + per[i][3]);
            per[i].push_back(per[i][1] + per[i][2] + per[i][3]);

            sort(per[i].begin(), per[i].end(), greater<int>());
        }

        _for(i, 0, n) rk.push_back(read() - 1);

        // then solve the problem
        int ans = solve();
        if(ans == -1) printf("Case %d: No solution\n", kase);
        else printf("Case %d: %d.%02d\n", kase, ans / 100, ans % 100);
    }
}