题目大意:有N个城市,给定计算两两城市距离的公式,然后求0到1~N-1的城市中,最短路径模掉M的最小值。
解题思路:先根据公式求出距离C矩阵,注意中间连乘3次的可能爆long long,然后用裸的dijstra算法求最短路。
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
const int maxd = 1005;
const int maxn = maxd * maxd;
const ll INF = 0x3f3f3f3f3f3f3f3f;
int N, M;
ll X[maxn], Y[maxn], Z[maxn];
ll C[maxd][maxd], d[maxd];
void init_distant () {
for (int i = 1; i < N * N; i++) {
if (i >= 2) {
X[i] = (12345 + X[i-1] * 23456 + X[i-2] * 34567 + (X[i-1] * X[i-2] % 5837501) * 45678) % 5837501;
Y[i] = (56789 + Y[i-1] * 67890 + Y[i-2] * 78901 + (Y[i-1] * Y[i-2] % 9860381) * 89012) % 9860381;
}
Z[i] = (X[i] * 90123 + Y[i]) % 8475871 + 1;
}
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++) {
if (i == j)
C[i][j] = 0;
else
C[i][j] = Z[i*N+j];
}
}
}
ll dijstra () {
int v[maxd];
memset(v, 0, sizeof(v));
for (int i = 0; i < N; i++)
d[i] = INF;
d[0] = 0;
ll ans = INF;
for (int i = 0; i < N; i++) {
int u = 0;
ll dis = INF;
for (int j = 0; j < N; j++) {
if (d[j] < dis && v[j] == 0) {
u = j;
dis = d[j];
}
}
v[u] = 1;
if (d[u] % M < ans && u)
ans = d[u] % M;
for (int j = 0; j < N; j++) {
if (d[u] + C[u][j] < d[j])
d[j] = d[u] + C[u][j];
}
}
return ans;
}
int main () {
while (scanf("%d%d%lld%lld%lld%lld", &N, &M, &X[0], &X[1], &Y[0], &Y[1]) == 6) {
init_distant();
printf("%lld\n", dijstra());
}
/*
while (cin >> N >> M >> X[0] >> X[1] >> Y[0] >> Y[1]) {
init_distant();
cout << dijstra() << endl;
}
*/
return 0;
}
hdu 4849 Wow! Such City!(dijstra)
时间: 2024-10-29 10:46:33