hdu 4849 Wow! Such City!(dijstra)

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题目链接：hdu 4849 Wow! Such City!

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;}

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