AtCoder Beginner Contest 211(补题)

C - chokudai

题意： 问字符串s有多少个子序列，是字符串t “chokudai”.
思路： 动态规划问题，设置 d p f [ i , j ] dpf[i,j] dpf[i,j]为取前长度为i的字符串s，与字符串t前长度为j匹配的数量。
那么此时,状态转移方程（官方题解，不想再手打了 ）

由于字符串t下标是从0开始的，所以下面的代码，方程转移会有小变化。
笔记：这个解法，适用于所有寻找字符串s中找有多少个子序列t(在不超时的情况下)

#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int N = 1e5 + 10, mod = 1e9 + 7;
char s1[N];
char s2[] = "chokudai";
int f[N][10];
int main() {
    scanf("%s", s1 + 1);
    int len = strlen(s1 + 1);
    f[0][0] = 1;
    for (int i = 1; i <= len; i++) {
        for (int j = 0; j <= 8; j++) {
            f[i][j] = f[i - 1][j];
        }
        for (int j = 0; j < 8; j++) {
            if (s2[j] == s1[i]) {
                f[i][j + 1] = (ll)f[i][j + 1] + f[i - 1][j] % mod;
            }
        }
    }
    printf("%d\n", f[len][8]);
    return 0;
}

D - Number of Shortest paths

题意： 问从1到N的最短路的路径数（裸模板题）
思路： 边权为1，基础最短路路径数的题，也可以开vector来存边

#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int N = 2 * 2e5 + 10, mod = 1e9 + 7;
queue<int> q;
int n, m;
int e[N], ne[N], h[N], idx;
int path[N], dis[N], vis[N];
void add(int a, int b) { e[idx] = b, ne[idx] = h[a], h[a] = idx++; }

int main() {
    scanf("%d%d", &n, &m);
    memset(h, -1, sizeof(h));
    for (int i = 0; i < m; i++) {
        int x, y;
        scanf("%d%d", &x, &y);
        add(x, y), add(y, x);
    }
    path[1] = 1;
    q.push(1);
    while (q.size()) {
        int t = q.front();
        q.pop();
        for (int i = h[t]; i != -1; i = ne[i]) {
            int j = e[i];
            if (!vis[j]) {
                vis[j] = 1;
                dis[j] = dis[t] + 1;
                q.push(j);
            }
            if (dis[j] == dis[t] + 1) {
                path[j] = (ll)(path[j] + path[t]) % mod;
            }
        }
    }
    printf("%d\n", path[n]);
    return 0;
}

#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int N = 1e6 + 10, mod = 1e9 + 7;
queue<int> q;
int n, m;
int e[N], ne[N], h[N], idx;
pair<int, int> f[N];
void add(int a, int b) { e[idx] = b, ne[idx] = h[a], h[a] = idx++; }

int main() {
    scanf("%d%d", &n, &m);
    memset(h, -1, sizeof(h));
    for (int i = 0; i < m; i++) {
        int x, y;
        scanf("%d%d", &x, &y);
        add(x, y), add(y, x);
    }
    f[1] = make_pair(0, 1);
    for (int i = 2; i <= n; i++) {
        f[i] = make_pair(1e9, 0);
    }
    q.push(1);
    while (!q.empty()) {
        int t = q.front();
        q.pop();
        for (int i = h[t]; i != -1; i = ne[i]) {
            int x = e[i];
            if (f[x].first > f[t].first + 1) {
                f[x] = f[t];
                f[x].first++;
                q.push(x);
            } else if (f[x].first == f[t].first + 1) {
                f[x].second = (ll)(f[x].second + f[t].second) % mod;
            }
        }
    }
    printf("%d\n", f[n].second);
    return 0;
}

To be continued
如果你有任何建议或者批评和补充，请留言指出，不胜感激

标签：AtCoder,idx,211,int,ll,add,补题,path,mod
来源： https://blog.csdn.net/weixin_54699118/article/details/119086013