pku 1797 Heavy Transportation 分支界限法

题目

http://acm.pku.edu.cn/JudgeOnline/problem?id=1797

 

分支界限法,学好了很强大.ps,第一次用优先队列...

 

#include <stdio.h>#include <queue>using namespace std;#define max 1001#define MIN(x,y)(x < y ? x : y)int map[max][max];int dist[max];//前驱节点int pre[max];int n,m;struct HeapNode{int i;//顶点int d;//从源点到i的当前路径上的最小权值HeapNode(int i = 0,int d = 0){this->i = i;this->d = d;}friend bool operator < (HeapNode a,HeapNode b){return a.d < b.d;}};//dist[i]:从源点到i的多条路径中每条路径上权值最小的边这一集合中的最大值//常规标记dijkstra法:dist[i] = max(dist[i],min(d[k],map[k][i]));//分支界限法:dist[i] = max(dist[i],min(enode(i == k).d,map[k][i]));void dijkstra(int u){memset(pre,0,sizeof(pre));memset(dist,0,sizeof(dist));HeapNode node,enode;priority_queue<HeapNode> p_q;enode.i = u;enode.d = 0;p_q.push(enode);//先多加几个节点进去是因为控制约束在根节点不起作用...for(int j = 1;j <= n;j++)if(map[enode.i][j]){node.i = j;node.d = map[enode.i][j];p_q.push(node);}while(!p_q.empty()){enode = p_q.top();p_q.pop();//搜索问题的解空间for(int j = 1;j <= n;j++)//控制约束:经过顶点i到顶点j的该条路径上的最小权值大于目前到顶点j的所有路径上最小值的最大值if(map[enode.i][j] && MIN(enode.d,map[enode.i][j]) > dist[j]){dist[j] = MIN(enode.d,map[enode.i][j]);pre[j] = enode.i;node.i = j;node.d = dist[j];p_q.push(node);}}}int main(){//freopen("data.txt","r",stdin);int _case,__case = 0;scanf("%d",&_case);while(++__case <= _case){memset(map,0,sizeof(map));scanf("%d%d",&n,&m);int i;int s,e,w;for(i = 1;i <= m;i++){scanf("%d%d%d",&s,&e,&w);map[s][e] = w;map[e][s] = w;}dijkstra(1);printf("Scenario #%d:/n%d/n/n",__case,dist[n]);}return 0;}