HDU ACM 1874 unblocked Project continued

Analysis: template problem, direct template can be.

#include <iostream> #include <queue>using namespace std;int u[2002];int v[2002];int w[2002];bool vis[202];    int D[202];int first[202];int next[2002];void Init (int n,int m) {int i;        for (i=0;i<n;i++) {vis[i]=false;    First[i]=-1; } for (i=0;i<m;i++) next[i]=-1;}    void SPFA (int n,int s)//flag indicates whether the current process is a reverse-established diagram or a start graph, false reverse, true start {queue<int> q;    int i,x,y;   for (i=0;i<n;i++) d[i]= (i==s)? 0:0x7fffffff;    Initialize yourself to yourself as 0, other to yourself equivalent to infinity Q.push (s);        while (!q.empty ()) {X=q.front ();        Q.pop ();        Vis[x]=false;            for (I=first[x];i!=-1;i=next[i]) {y=v[i];                if (D[y]>d[x]+w[i]) {d[y]=d[x]+w[i];                             if (!vis[y]) {vis[y]=true;                Q.push (y);    }}}}}void Read (int m) {int i,a,b; for (i=0;i<m;i++) {scanf ("%d%d%d", &a,&b,& w[i]);                        U[i]=a;        Store forward edge, because it is an v[i]=b graph;        Next[i]=first[a];        First[a]=i;                W[i+1]=w[i];        Store reverse Edge i++;        U[i]=b;        V[i]=a;        NEXT[I]=FIRST[B];    First[b]=i; }}void Output (int t) {if (D[T]==0X7FFFFFFF) puts ("1"); elseprintf ("%d\n", D[t]);}    int main () {int n,m;    int s,t;  while (scanf ("%d%d", &n,&m) ==2) {Init (n,m+m);        M+m is because there is no edge, each to save two Read (m+m);        scanf ("%d%d", &s,&t);        SPFA (n,s);    Output (t);    } return 0; }

HDU ACM 1874 unblocked Project continued