Luogu P1144 最短路计数题解

题目大意

给你一个\(n\)个点的无向图和\(m\)条边的有向图，然后我们需要求出从\(1\)号点到达图中每个点的最短路径有多少条。
( \(n \le 1000000,m \le 2000000\))

题目传送门

Luogu P1144

思路

首先我们先用Bfs求出1号点到达每个点的最短路（因为题目中说了每条边权值为1）用一个\(min_dis\)数组来存储从1号点到达每个点的最小值。然后我们再次进行bfs，然后我们设\(f[i]\)为从\(1\)到达结点\(i\)的最短路数量。那么我们在第二次bfs的过程中，对于当前结点\(x\)，我们判断所有与之相连的点\(v\)是否满足\(min_dis[x] = min_dis[v] - 1\)。如果满足就说明这条边一定在从\(1\)到\(v\)的最短路上那么说明\(f[v]\)的值可以从\(f[x]\)转移过来，然后我们将\(f[v]\)加上\(f[x]\)即可。注意初始化\(f[1]\)为\(1\)。

代码：

#include 
#include 
#include 
#include 
#include 

using namespace std;

const int N = 1e6 + 5;
const int M = 4e6 + 5;
const int MOD = 100003;

int n, m;
int u, v;
int cnt = 0;
int head[N];
int f[N];
int min_dis[N];
bool vis[N];

struct EDGE {
    int s;
    int e;
    int nxt;
} Edge[M];

inline int Min(int x, int y) {
    return x <= y ? x : y;
}

int read() { //快速读入函数
    int s = 0, w = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
        if (ch == '-') w = -1;  
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
        s = s * 10 + ch - '0';
        ch = getchar();
    }
    return s * w;
}

void write(int x) { //快速输出函数
    if (x < 0) putchar('-'), x = -x;
    if (x > 9) write(x / 10);
    putchar(x % 10 + '0');
}

void add(int u, int v) {
    ++cnt;
    Edge[cnt].s = u;
    Edge[cnt].e = v;
    Edge[cnt].nxt = head[u];
    head[u] = cnt;
}

void bfs(int x) { //第一次宽搜的函数
    queue  q;
    q.push(x);
    vis[x] = true;
    while (!q.empty()) {
        int now = q.front(); q.pop();
        for (register int i = head[now]; i; i = Edge[i].nxt) {
            if (vis[Edge[i].e]) continue;
            vis[Edge[i].e] = true;
            min_dis[Edge[i].e] = min_dis[now] + 1; //求最短路径
            q.push(Edge[i].e);
        }
    }
}

void bfs1(int x) { //第二次宽搜的函数
    queue  q;
    q.push(x);
    vis[x] = true;
    while (!q.empty()) {
        int now = q.front(); q.pop();
        for (register int i = head[now]; i; i = Edge[i].nxt) {
            if (vis[Edge[i].e]) continue;
            vis[Edge[i].e] = true;
            q.push(Edge[i].e);
        }
        for (register int i = head[now]; i; i = Edge[i].nxt) {
            if (min_dis[now] == min_dis[Edge[i].e] - 1) {
                f[Edge[i].e] = (f[now] % MOD + f[Edge[i].e] % MOD) % MOD; //递推求出答案
            }
        }
    }
}

int main(int argc, char const *argv[]) {
    n = read(), m = read();
    for (register int i = 1; i <= m; ++i) {
        u = read(), v = read();
        add(u, v);
        add(v, u); //建立双向边
    }
    bfs(1);
    memset(vis, 0, sizeof(vis)); //初始化vis数组
    f[1] = 1;
    bfs1(1);
    for (register int i = 1; i <= n; ++i) 
        write(f[i]), putchar('\n');
    return 0;
}

转载于:https://www.cnblogs.com/lixiao189/p/9858399.html