杭电 3790 最短路径问题
题目地址:
http://acm.hdu.edu.cn/showproblem.php?pid=3790
没什么太大的感觉,只是觉得进步是一个过于缓慢的感觉……
dijkstra算法的应用,注意可重边的处理以及同等权值的路径时,路径的选择
#include <stdio.h> #define MAXN 1010 #define INF 1000000000 int mat[MAXN][MAXN]; int bb[MAXN][MAXN]; int min[MAXN]; int pre[MAXN]; int com(int m, int s, int a, int b) { int path1[MAXN], path2[MAXN], i = 2, j = 2; int sum1 = 0, sum2 =0, p; path1[0] = path2[0] = m; path1[1] = a; path2[1] = b; while( a != s ) { path1[i++] = pre[a]; a = pre[a]; } while( b != s ) { path2[j++] = pre[b]; b = pre[b]; } for( p = 0; p < (i-1); p++ ) sum1 += bb[path1[p]][path1[p+1]]; for( p = 0; p < (j-1); p++ ) sum2 += bb[path2[p]][path2[p+1]]; if( sum1 >= sum2 ) return path2[1]; return path1[1]; } void Dijkstra(int n,int s) { int v[MAXN], i, j, k; for( i = 1; i <= n; i++ ) v[i] = 0, min[i] = INF, pre[i] = -1; for( min[s] = 0, j = 1; j <= n; j++ ) { for( k = 0, i = 1; i <= n; i++ ) if( (!v[i]) && ((k == 0)||(min[i] < min[k])) ) k = i; for( v[k] = 1, i = 1; i <= n; i++ ) if( (!v[i]) && (mat[k][i] != INF) && (mat[k][i] + min[k] < min[i]) ) { min[i] = min[k] + mat[k][i]; pre[i] = k; } else if( !v[i] && (mat[k][i] != INF) && (mat[k][i] + min[k] == min[i]) ) pre[i] = com(i,s,pre[i],k); } } int main() { int n, m, k, s, t, sum, end; int i, j, a, b; while( (scanf( "%d%d", &n, &m)!= EOF) && n && m ) { for( i = 1; i <= n; i++ ) for( j = 1; j <= n; j++ ) { mat[i][j] = INF; bb[i][j] = INF; } for( k = 0; k < m; k++ ) { scanf("%d%d%d%d", &i, &j, &a, &b); if( a < mat[i][j] ) { mat[j][i] = mat[i][j] = a; bb[j][i] = bb[i][j] = b; } else if( (a == mat[i][j])&&(b < bb[i][j]) ) bb[i][j] = bb[j][i] = b; } /*for( i = 1; i <= n; i++ ) { for( j = 1; j <= n; j++ ) printf( "%d ", mat[i][j] ); printf( "\n" ); }*/ scanf("%d%d", &s, &t); //printf("%d %d\n",s,t); Dijkstra(n,s); /*for( i = 1; i <= n; i++ ) printf( "%d ", min[i] ); printf( "\n" ); for( i = 1; i <= n; i++ ) printf( "%d ", pre[i] ); printf( "\n" );*/ //printf("123"); sum = 0; end = t; while( end != s ) { sum += bb[end][pre[end]]; end = pre[end]; } //printf("123"); printf("%d %d\n",min[t],sum); } return 0; }