题目大意：就是给定起始点和结束点以及卡车的重量，问你在保证卡车重量最大的前提下，最短路径是多少，我们可以枚举卡车的重量，在对每个卡车的重量进行dijkstra，算到终点是否存在最短路径。当然枚举也可以用二分进行优化，时间差的还是蛮多的。
| 2016-08-05 15:23:00 | Accepted | 2962 | 9297MS | 9500K | 2257 B | G++ |
| 2016-08-05 15:34:54 | Accepted | 2962 | 1606MS | 9496K | 2516 B | G++ |
下面那个是进行过二分优化的。
//二分+dijkstra
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define INF 0x3f3f3f3f
int n,m,ch,h,l,c,r,s,e,ans1,ans2;
bool visited[1005];
int a[1005][1005],height[1005][1005];
int dist[1005];
int hei[1005];
void dijkstra(int he)
{
memset(visited,false,sizeof(visited));
memset(dist,0x3f,sizeof(dist));
visited[s]=true;
for(int i=1;i<=n;i++)
{
hei[i]=height[s][i];
if(hei[i]<he)
dist[i]=INF;
else
dist[i]=a[s][i];
}
for(int i=1;i<=n;i++)
{
int MIN=INF,node=-1;
for(int j=1;j<=n;j++)
{
if(!visited[j]&&dist[j]<MIN&&hei[j]>=he)
{
MIN=dist[j];
node=j;
}
}
if(node==-1)
return;
visited[node]=true;
for(int j=1;j<=n;j++)
{
if(!visited[j]&&height[node][j]>=he&&dist[j]>dist[node]+a[node][j])
{
dist[j]=dist[node]+a[node][j];
hei[j]=height[node][j];
}
}
}
}
int main()
{
int sym=0;
while(true)
{
sym++;
ans1=-1;
ans2=-1;
memset(a,0x3f,sizeof(a));
memset(height,0,sizeof(height));
scanf("%d%d",&n,&m);
if(n==0&&m==0)
break;
for(int i=0;i<m;i++)
{
scanf("%d%d%d%d",&l,&r,&h,&c);
a[l][r]=a[r][l]=c;
if(h==-1)
height[l][r]=height[r][l]=INF;
else
height[l][r]=height[r][l]=h;
}
for(int i=1;i<=n;i++)
{
a[i][i]=height[i][i]=0;
}
scanf("%d%d%d",&s,&e,&ch);
int left=1;
int right=ch;
int mid=0;
while(left<=right)
{
mid=(left+right)/2;
dijkstra(mid);
if(dist[e]!=INF)
{
ans1=dist[e];
ans2=mid;
left=mid+1;
}
else
right=mid-1;
}
if(sym!=1)
printf("\n");
printf("Case %d:\n",sym);
if(ans1==-1&&ans2==-1)
{
printf("cannot reach destination\n");
}
else
{
printf("maximum height = %d\n",ans2);
printf("length of shortest route = %d\n",ans1);
}
}
return 0;
}
//枚举+dijkstra
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define INF 0x3f3f3f3f
int n,m,ch,h,l,c,r,s,e;
bool visited[1005];
int a[1005][1005],height[1005][1005];
int dist[1005];
int hei[1005];
void dijkstra(int he)
{
memset(visited,false,sizeof(visited));
memset(dist,0x3f,sizeof(dist));
visited[s]=true;
for(int i=1;i<=n;i++)
{
hei[i]=height[s][i];
if(hei[i]<he)
dist[i]=INF;
else
dist[i]=a[s][i];
}
for(int i=1;i<=n;i++)
{
int MIN=INF,node=-1;
for(int j=1;j<=n;j++)
{
if(!visited[j]&&dist[j]<MIN&&hei[j]>=he)
{
MIN=dist[j];
node=j;
}
}
if(node==-1)
return;
visited[node]=true;
for(int j=1;j<=n;j++)
{
if(!visited[j]&&height[node][j]>=he&&dist[j]>dist[node]+a[node][j])
{
dist[j]=dist[node]+a[node][j];
hei[j]=height[node][j];
}
}
}
}
int main()
{
int sym=0;
while(true)
{
sym++;
memset(a,0x3f,sizeof(a));
memset(height,0,sizeof(height));
scanf("%d%d",&n,&m);
if(n==0&&m==0)
break;
for(int i=0;i<m;i++)
{
scanf("%d%d%d%d",&l,&r,&h,&c);
a[l][r]=a[r][l]=c;
if(h==-1)
height[l][r]=height[r][l]=INF;
else
height[l][r]=height[r][l]=h;
}
for(int i=1;i<=n;i++)
{
a[i][i]=height[i][i]=0;
}
scanf("%d%d%d",&s,&e,&ch);
while(ch!=0)
{
dijkstra(ch);
if(dist[e]!=INF)
{
break;
}
ch--;
}
if(sym!=1)
printf("\n");
printf("Case %d:\n",sym);
if(ch==0)
{
printf("cannot reach destination\n");
}
else
{
printf("maximum height = %d\n",ch);
printf("length of shortest route = %d\n",dist[e]);
}
}
return 0;
}