2024-10-15 19:39阅读: 2评论: 0推荐: 0

20241010总结

number

从考虑使用多少个加和减入手，简单计算。

#include<iostream>
#define int long long

using namespace std;

inline int read(){register int x = 0, f = 1;register char c = getchar();while (c < '0' || c > '9'){if (c == '-') f = -1;c = getchar();}while (c >= '0' && c <= '9'){x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return x * f;}
inline void write(int x){if (x < 0) putchar('-'), x = -x;if (x > 9) write(x / 10);putchar(x % 10 + '0');}

int T, n, a, b, s;
int get(int x, int y){
    if (x % y == 0) return x / y;
    return x / y + 1;
}

signed main(){
    freopen("number.in", "r", stdin);
    freopen("number.out", "w", stdout);
    T = read();
    while (T--){
        n = read(), a = read(), b = read(), s = read();
        int x = s - n * b;
        if (x >= 0){
            cout << x << '\n';
            continue;
        }
        cout << x + get(-x, a + b) * (a + b) << '\n';
    }
    return 0;
}

path

首先可以想到一个 $d p$ ，设 $d p_{i}$ 表示在 $i$ 点时的答案，于是转移从所有与它连边的点转移， $d p_{u} = min_{u \to v} (d p_{u}, d p_{v} + [b_{d p_{u} + 1} = v_{v}])$ ，发现这个可以用 $d i j k s t r a$ 类似的算法实现，于是用堆对 $d p$ 进行优化。

#include<iostream>
#include<queue>
#include<vector>
#define int long long

using namespace std;

inline int read(){register int x = 0, f = 1;register char c = getchar();while (c < '0' || c > '9'){if (c == '-') f = -1;c = getchar();}while (c >= '0' && c <= '9'){x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return x * f;}
inline void write(int x){if (x < 0) putchar('-'), x = -x;if (x > 9) write(x / 10);putchar(x % 10 + '0');}

const int N = 5e5 + 10, M = 1e6 + 10;
int n, m, k;
int val[N], b[N];
struct edge{
    int v, nxt;
}e[M << 1];
int head[N], cnt;
void add(int u, int v){
    e[++cnt] = (edge){v, head[u]};
    head[u] = cnt;
}
int dp[N];
struct node{
    int d, u;
    bool operator < (const node &b) const{
        return d > b.d;
    }
};
priority_queue<node> q;
void bfs(int s){
    q.push((node){dp[s], s});
    while (!q.empty()){
        node t = q.top();q.pop();
        int u = t.u;
        if (t.d != dp[u]) continue;
        for (int i = head[u]; i; i = e[i].nxt){
            int v = e[i].v;
            if (dp[v] > dp[u] + (b[dp[u] + 1] == val[v])){
                dp[v] = dp[u] + (b[dp[u] + 1] == val[v]);
                q.push((node){dp[v], v});
            }
        }
    }
}

signed main(){
    freopen("path.in", "r", stdin);
    freopen("path.out", "w", stdout);
    n = read(), m = read(), k = read();
    for (int i = 1; i <= n; i++) val[i] = read();
    for (int i = 1; i <= k; i++) b[i] = read();
    for (int i = 1; i <= m; i++){
        int u = read(), v = read();
        add(u, v), add(v, u);
    }
    for (int i = 1; i <= n; i++) dp[i] = 0x7fffffff;
    dp[1] = (val[1] == b[1] ? 1 : 0);
    bfs(1);
    for (int i = 1; i <= n; i++) cout << dp[i] << ' ';
    return 0;
}