10/17 NOIP模拟赛

期望得分：50；实际得分：0；

考场思路：暴力（完全没有往贪心的方面想，还差点把ssj带沟里 qwq）

拿到数据发现，这样写暴力完全不对啊。。。。

很多情况没有考虑到 qwq

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
const int M = 1005;
int n, len, sum;
int p, q, maxn;
char s[M][M];

int main() {
//    freopen("curse.in","r",stdin);
//    freopen("curse.out","w",stdout);
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) {
        cin >> s[i];
        len = strlen(s[i]);
    }
    for (int i = 1; i <= n; ++i)
        for (int j = i + 1; j <= n; ++j) {
            for (int k = 0; k < len; ++k)
                if (s[i][k] == s[j][k]) ++sum;
            if (sum > maxn) {
                maxn = sum;
                p = i, q = j;
            }
            sum = 0;
        }
    int y = 0, x = 0;
    for (int i = 0; i < len; ++i) {
        if (s[p][i] == '0') ++x;
        if (s[q][i] == '0') ++y;
    }
    if (x > y) cout << s[p] << '\n';
    else cout << s[q] << '\n';
//    fclose(stdin); fclose(stdout);
    return 0;
}

考场代码

正解：贪心，比较每一位上 0、1 出现的次数，若 1 出现次数 > n/2 输出 1，反之输出 0

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;

int n, len;
int b[1005];
string s;

int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) {
        cin >> s;
        len = s.length();
        for (int j = 0; j < len; ++j) {
            if (s[j] == '1') ++b[j];
        }
    }
    for (int i = 0; i < len; ++i)
        if (b[i] > n / 2) printf("1");
        else printf("0");
    return 0;
}

View Code

10/17 NOIP模拟赛

考场思路：二分+？

好像与处理很麻烦的样子。。。不会要用DP吧。。。

然后就放弃了。。。

正解：二分+DP

特判一下，当 R + G ≥ N 时，L = 1 即可全部覆盖，所以DP的范围便从 10⁹ 减小到 2 * 10³

将祭坛的位置按从小到大排序，然后二分求 L 的大小

预处理 P、Q 两个数组，记录在祭坛 i 使用法杖能摧毁的最远的祭坛

用DP判断二分的 L 是否符合题目要求

#include <algorithm>
#include <cstring>
#include <cstdio>
using namespace std;

const int M = 2005;
int n, p, q, a[M];
int l = 1, r, ans;
int dp[M][M], P[M], Q[M];

bool check(int L) {
    memset(dp, 0, sizeof dp);
    memset(P, 0, sizeof P);
    memset(Q, 0, sizeof Q);
    for (int i = 1; i <= n; ++i)  //预处理 
        for (int j = i; j <= n; ++j) {
            if (a[j] - a[i] + 1 <= L) P[i] = j;
            if (a[j] - a[i] + 1 <= 2 * L) Q[i] = j;
        }
//    P[n + 1] = Q[n + 1] = n;  与第25行意义相同，防止DP时超过边界 
    for (int i = 0; i <= p; ++i)
        for (int j = 0; j <= q; ++j) {
            if (i > 0) dp[i][j] = max(dp[i][j], P[dp[i - 1][j] + 1]);
            if (j > 0) dp[i][j] = max(dp[i][j], Q[dp[i][j - 1] + 1]);
            if (dp[i][j] == n) return true;
        }
    return dp[p][q] == n ? true : false;
}

int main() {
//    freopen("light.in", "r", stdin);
//    freopen("light.out", "w", stdout);
    scanf("%d%d%d", &n, &p, &q);
    for (int i = 1; i <= n; ++i)
        scanf("%d", &a[i]);
    sort(a + 1, a + 1 + n);
    a[0] = 0;
    r = a[n] - a[1] + 1;
    if (p + q >= n) return printf("1\n"), 0;
    while (l <= r) {  //二分 
        int mid = (l + r) / 2;
        if (check(mid)) ans = mid, r = mid -1;
        else l = mid + 1;
    }
    printf("%d\n", ans);
    return 0;
}

View Code

10/17 NOIP模拟赛

思路：求严格次短路

#include <cassert>
#include <cstring>
#include <cstdio>
#include <queue>
using namespace std;

#define LOOP(i,n) for(int i=1; i<=n; i++)
const int MAX_NODE = 5010;
const int MAX_EDGE = 100010 << 1;
const int INF = 0x3f3f3f3f;

struct Node;
struct Edge;

struct Node {
    int Id, Dist, Dist2;
    bool Inq;
    Edge *Head;
} _nodes[MAX_NODE], *Start, *Target;
int _vCount;

struct Edge {
    int Weight;
    Node *From, *To;
    Edge *Next;
    Edge() {}
    Edge(Node *from, Node *to, Edge *next, int weight) :
        From(from), To(to), Next(next), Weight(weight) {}
}*_edges[MAX_EDGE];
int _eCount;

void init(int vCount) {
    memset(_nodes, 0, sizeof(_nodes));
    _vCount = vCount;
    _eCount = 0;
    Start = 1 + _nodes;
    Target = vCount + _nodes;
}

void AddEdge(Node *from, Node *to, int weight) {
    Edge *e = _edges[++_eCount] = new Edge(from, to, from->Head, weight);
    e->From->Head = e;
}

void Build(int uId, int vId, int weight) {
    Node *u = uId + _nodes, *v = vId + _nodes;
    u->Id = uId;
    v->Id = vId;
    AddEdge(u, v, weight);
    AddEdge(v, u, weight);
}

void SPFA() {
    LOOP(i, _vCount)
    _nodes[i].Dist = _nodes[i].Dist2 = INF;
    static queue<Node*> q;
    Start->Dist = 0;
    Start->Dist2 = INF;
    Start->Inq = true;
    q.push(Start);
    while (!q.empty()) {
        Node *u = q.front();
        q.pop();
        u->Inq = false;
        for (Edge *e = u->Head; e; e = e->Next) {
            bool relaxOk = false;
            if (u->Dist + e->Weight < e->To->Dist) {
                e->To->Dist2 = e->To->Dist;
                e->To->Dist = u->Dist + e->Weight;
                relaxOk = true;
            }
            else if (u->Dist + e->Weight > e->To->Dist && u->Dist + e->Weight < e->To->Dist2) {
                e->To->Dist2 = u->Dist + e->Weight;
                relaxOk = true;
            }
            if (u->Dist2 + e->Weight < e->To->Dist2) {
                e->To->Dist2 = u->Dist2 + e->Weight;
                relaxOk = true;
            }
            if (relaxOk && !e->To->Inq) {
                e->To->Inq = true;
                q.push(e->To);
            }
        }
    }
}

int main() {
//    freopen("maze.in","r",stdin);
//    freopen("maze.out","w",stdout);
    int testCase, totNode, totEdge, uId, vId, weight, sId, tId;
    scanf("%d%d", &totNode, &totEdge);
    init(totNode);
    LOOP(i, totEdge) {
        scanf("%d%d%d", &uId, &vId, &weight);
        Build(uId, vId, weight);
    }
    SPFA();
    printf("%d\n", Target->Dist2);
//    fclose(stdin); fclose(stdout);
    return 0;
}

View Code

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10/17 NOIP模拟赛

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