The graph of data structure experiment seven: The best short Circuit of the Donkey Friend program minimum cost __ Shortest

graph of data structure experiment seven: The Donkey friend plan Time limit:1000ms Memory limit:65536k Topic DescriptionAs a senior donkey friend, small new has a collection of self-driving tour map, the figure detailed on the national highway between the city distance and highway charges, now you write a program to find a way to the shortest path between the destination, if there are more than one path shortest, then the output of the least travel path. input

Continuous T-Group data entry, the first row of each set of input data gives four positive integer n,m,s,d, where n (2 <= n <= 500) is the number of cities, the city number from 0~n-1,m is the number of urban expressway, S is the city number of departure, D is the city number of the destination; then M line  , each line gives a highway information, said City 1, City 2, highway length, charge amount, the middle by space interval, the number is integer and not more than 500, the input data are guaranteed to have the solution. Output

Output path length and total charges in the same row, with space spacing between the data. Sample Input

1
4 5 0 3 0 1 1 1 3 2 0 3 4 0 2 2 2

Sample Output

3 40

Tips

#include <bits/stdc++.h> using namespace std;
const int inf=20000000;
const int maxn=1000;
int MAP[MAXN][MAXN];
int CON[MAXN][MAXN];
int n,m,s,d; int Floyd () {for (int i=0;i<n;i++) {for (int j=0;j<n;j++) {for (int k=0;k<n;k + +) {if ((map[j][k]>map[j][i]+map[i][k)) | |
                (Map[j][k]==map[j][i]+map[i][k]&&con[j][k]>con[j][i]+con[i][k])
                    {Map[j][k]=map[j][i]+map[i][k];
                CON[J][K]=CON[J][I]+CON[I][K];
}} printf ("%d%d\n", map[s][d],con[s][d]);
    int main () {int T;
            while (~SCANF ("%d", &t)) {while (t--) {scanf ("%d%d%d%d", &n,&m,&s,&d);
                    for (int i=0;i<=n;i++) {for (int j=0;j<=n;j++) {
                    if (i==j) map[i][j]=con[i][j]=0;
          Else          Map[i][j]=con[i][j]=inf;
                } for (int i=0;i<m;i++) {int u,v,w,x;
                scanf ("%d%d%d%d", &u,&v,&w,&x);
                     if (map[u][v]>w) {map[u][v]=map[v][u]=w;
                    Con[u][v]=con[v][u]=x;
        } Floyd ();
} return 0;
 }

#include <bits/stdc++.h> using namespace std;
const int inf=100000000;
const int maxn=1000000;
int n,m,s,d,lsum=0,msum=0;
struct node {int l,m,u,v;} EDGE[MAXN];
int DIS[MAXN],CON[MAXN];
        int Bellman () {for (int i=0; i<=n; i++) {dis[i]=inf;
    Con[i]=inf;
    } dis[s]=0;
    con[s]=0; for (int i=1; i<n; i++) {for (int j=0; j<2*m; J + +) {if (edge[j). L+DIS[EDGE[J].U]&LT;DIS[EDGE[J].V]) | | ((EDGE[J). L+DIS[EDGE[J].U]==DIS[EDGE[J].V]) && (Edge[j]. M+CON[EDGE[J].U]&LT;CON[EDGE[J].V]))) {Dis[edge[j].v]=edge[j].
                L+DIS[EDGE[J].U]; CON[EDGE[J].V]=CON[EDGE[J].U]+EDGE[J].
            M
} printf ("%d%d\n", dis[d],con[d]);
    int main () {int t,v,u,w,x;
            while (~SCANF ("%d", &t)) {while (t--) {scanf ("%d%d%d%d", &n,&m,&s,&d);
            int num=0;
    for (int i=0; i<m; i++) {            scanf ("%d%d%d%d", &u,&v,&w,&x); Edge[num].
                L=w; Edge[num].
                M=x;
                Edge[num].u=u;
                 Edge[num++].v=v; Edge[num].
                L=w; Edge[num].
                M=x;
                Edge[num].v=u;
            Edge[num++].u=v;
        } bellman ();
} return 0;
 }