HDU - 6026 D - Deleting Edges
作者:互联网
最短路计数
https://vjudge.net/contest/464901#problem/D
题意是给一个图,问最后这个图使得每个点到 \(1\) 点距离都是原图中最短的生成树的个数。
虽然说很明显只要求最短路个数就行了,但是我也不懂为啥最短路计数是这样记的。
题解都说是最短路计数的模板...话说咱也不知道为啥啊。答案是将每个点的贡献乘起来。
需要再理解下。。
点击查看代码
#include <algorithm>
#include <cstring>
#include <iostream>
#include <list>
#include <map>
#include <queue>
#include <set>
#include <vector>
using namespace std;
typedef long long ll;
#define P pair<int, int>
const int N = 50 + 5;
const ll INF = 1e18;
const int mod = 1e9 + 7;
int n;
ll mp[N][N], dis[N], vis[N];
ll cnt[N];
void dij() {
priority_queue<P, vector<P>, greater<P> > q;
memset(dis, 0x3f, sizeof dis);
memset(vis, 0, sizeof vis);
memset(cnt, 0, sizeof cnt);
q.push(P{0, 1});
dis[1] = 0;
while (q.size()) {
P now = q.top();
q.pop();
int o = now.second, d = now.first;
if (vis[o]) continue;
vis[o] = 1;
for (int i = 1; i <= n; i++) {
if (i == o || mp[o][i] == 0) continue;
if (dis[i] > dis[o] + mp[o][i]) {
dis[i] = dis[o] + mp[o][i];
q.push(P{dis[i], i});
cnt[i] = 1;
} else if (dis[i] == dis[o] + mp[o][i]) {
cnt[i]++;
}
}
}
}
string str;
int main() {
while (cin >> n) {
for (int i = 1; i <= n; i++) {
cin >> str;
for (int j = 1; j <= n; j++) {
mp[i][j] = str[j-1] - '0';
if (mp[i][j] == 0 && i != j) mp[i][j] = INF;
}
}
// cout << "---\n";
dij();
ll ans = 1;
for (int i = 2; i <= n; i++) {
if (dis[i] == INF) ans = 0;
ans *= cnt[i];
ans %= mod;
}
cout << ans << endl;
}
}
标签:HDU,int,ll,cnt,vis,Edges,include,Deleting,dis 来源: https://www.cnblogs.com/FushimiYuki/p/15484374.html