【PAT甲级】1072. Gas Station (30)

此题用Dijkstra计算；如果用Floyd最后一个测试用例会超时

#include <stdio.h>#include <iostream> #include <vector>#include <string>#include <algorithm>#include <stdlib.h>#define INF 1<<29using namespace std;struct T {    int min;    int sum;    int index;};int min(int a, int b); T dij(int start);bool cmp(T a, T b);int n, m, k, ds;int N;int cost[1100][1100];int main() {    cin >> n >> m >> k >> ds;    N = m + n;    int p1, p2, d;    string p1str, p2str;    vector<T> v;    for (int i = 0; i <= N; i++) {        for (int j = 0; j <= N; j++) {            if (i == j) cost[i][j] = 0;            else cost[i][j] = INF;        }    }    for (int i = 0; i < k; i++) {        cin >> p1str >> p2str >> d;        if (p1str[0] == 'G') p1 = n + atoi(p1str.substr(1).c_str());        else p1 = atoi(p1str.c_str());        if (p2str[0] == 'G') p2 = n + atoi(p2str.substr(1).c_str());        else p2 = atoi(p2str.c_str());        cost[p1][p2] = cost[p2][p1] = d;    }    for (int i = n + 1; i <= N; i++) {        T t = dij(i);        if (t.min != -1) v.push_back(t);    }    sort(v.begin(), v.end(), cmp);    if (v.size())         printf("G%d\n%.1f %.1f\n", v[0].index - n, float(v[0].min), float(v[0].sum)/n);    else         printf("No Solution");    return 0;}int min(int a, int b) {    return (a < b ? a : b);}bool cmp(T a, T b) {    if (a.min > b.min) return true;    else if (a.min == b.min && a.sum < b.sum) return true;    else if (a.min == b.min && a.sum == b.sum && a.index < b.index) return true;    else return false;}T dij(int start) {    int *d = new int[N+1];    int *used = new int[N+1];    fill(d, d + N + 1, INF);    fill(used, used + N + 1, 0);    d[start] = 0;    while (true) {        int v = -1;        for (int u = 1; u <= N; u++) {            if (!used[u] && (v == -1 || d[u] < d[v]))                v = u;        }        if (v == -1) break;        used[v] = 1;        for (int u = 1; u <= N; u++) {            d[u] = min(d[u], d[v] + cost[v][u]);        }    }    int min = INF;    int sum = 0;    for (int i = 1; i <= n; i++) {        if (d[i] < min) min = d[i];        sum += d[i];    }    for (int i = 1; i <= n; i++) {        if (d[i] > ds) min = -1;    }    T t;    t.min = min;    t.sum = sum;    t.index = start;    return t;}