CSU-1307-CityTour+Dij+Kruskal

/*最短路+最小生成树题意：给定一张图，起点，终点。求起点到终点的一条路（这条路经过的最长的一段要最短！）枚举这条“最长的路”，可二分，也可直接计算出。*/#include<stdio.h>#include<string.h>#include<stdlib.h>#include<algorithm>#include<queue>using namespace std;const int maxn = 2005;const int maxm = 50005;const int inf = 99999999;int mat[ maxn ][ maxn ];int fa[ maxn ];bool vis[ maxn ];int dis[ maxn ];struct Node{int x,y,val;}edge[ maxm<<1 ];int cmp( Node a,Node b ){return a.val<b.val;}void init( int n ){for( int i=1;i<=n;i++ ){fa[ i ] = i;for( int j=1;j<=n;j++ ){mat[i][j] = inf;}}}int find( int x ){if( x==fa[x] ) return x;return fa[x] = find(fa[x]);}int Kruskal( int n,int m,int A,int B ){sort( edge,edge+m,cmp );int x,y;int MaxEdge = 0;for( int i=0;i<m;i++ ){x = find( edge[i].x );y = find( edge[i].y );if( x!=y ){if( x>y ) fa[y] = x;else fa[x] = y;if( find(A)==find(B) ){MaxEdge = edge[i].val;return MaxEdge;}MaxEdge = max( MaxEdge,edge[i].val );}}return MaxEdge;}int Dij( int n,int MaxEdge,int A,int B ){for( int i=1;i<=n;i++ ){vis[i] = false;dis[i] = inf;}dis[ A ] = 0;for( int i=1;i<=n;i++ ){int M = inf;int id = -1;for( int j=1;j<=n;j++ ){if( !vis[j] && M>dis[j] ){M = dis[j];id = j;}}if( id==-1 ) break;vis[ id ] = true;for( int j=1;j<=n;j++ ){if( !vis[j] && dis[j]>dis[id]+mat[id][j] && mat[id][j]<=MaxEdge ){dis[j] = dis[id]+mat[id][j];}}}return dis[B];}int main(){//freopen("in.txt","r",stdin);int n,m,A,B;while( scanf("%d%d%d%d",&n,&m,&A,&B)!=EOF ){init( n );for( int i=0;i<m;i++ ){scanf("%d%d%d",&edge[i].x,&edge[i].y,&edge[i].val);if( mat[edge[i].x][edge[i].y]>edge[i].val ){mat[edge[i].x][edge[i].y] = mat[edge[i].y][edge[i].x] = edge[i].val;}}int MaxEdge = 0;MaxEdge = Kruskal( n,m,A,B );//printf("MaxEdge = %d\n",MaxEdge);int Sum = Dij( n,MaxEdge,A,B );printf("%d\n",Sum);}}