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