UVA1416 Warfare And Logistics(LA4080)（最短路，5级）
Description
The army of United Nations launched a new wave of air strikes on terrorist forces. The objective of the mission is to reduce enemy's logistical mobility. Each air strike will destroy a path and therefore increase the shipping cost of the shortest path between two enemy locations. The maximal damage is always desirable.
Let's assume that there are n enemy locations connected by m bidirectional paths, each with specific shipping cost. Enemy's total shipping cost is given as c = path(i, j) . Here path(i, j) is the shortest path between locations i and j . In case i and j are not connected, path(i, j) = L . Each air strike can only destroy one path. The total shipping cost after the strike is noted as c' . In order to maximized the damage to the enemy, UN's air force try to find the maximal c' - c .
Input
The first line ofeach input case consists ofthree integers: n , m , and L . 1 < n100 , 1m1000 , 1L108 . Each ofthe followingm lines contains three integers: a , b , s , indicating length of the path between a and b .
Output
For each case, output the total shipping cost before the air strike and the maximal total shipping cost after the strike. Output them in one line separated by a space.
Sample Input
4 6 1000 1 3 2 1 4 4 2 1 3 2 3 3 3 4 1 4 2 2
Sample Output
28 38
思路：先进行n次最短路，算出原先的和sum_all;
然后对于删除每一条边所影响的最短路，进行重新最短路，算出新的sum_del,
900MS+
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<queue>
using namespace std;
typedef long long LL;
const LL oo=1e18;
const int mm=100+9;
const int nn=2000+9;
class Edge
{
public:int v,next;LL dis;
}e[nn];
class heap
{
public:int u;LL dis;
bool operator <(const heap& x)const
{
return dis>x.dis;
}
};
int head[mm],edge,n,m,L;
bool del[nn],vis[mm],belong[nn][mm];LL dis[mm][mm],dd[mm],sum_all,sum_del,sum_mid;
void data()
{
memset(head,-1,sizeof(head));edge=0;
}
void add(int u,int v,LL _dis)
{
e[edge].v=v;e[edge].dis=_dis;e[edge].next=head[u];head[u]=edge++;
}
void dijstra(LL*d,int S,bool action)
{
memset(vis,0,sizeof(vis));
priority_queue<heap>Q;
for(int i=0;i<=n;++i)
d[i]=oo;
d[S]=0;
Q.push((heap){S,0});
heap z;int u,v;
while(!Q.empty())
{
z=Q.top();Q.pop();
u=z.u;
if(vis[u])continue;vis[u]=1;
for(int i=head[u];~i;i=e[i].next)
{ v=e[i].v;
if(action&&del[i])continue;///i边已经删除
if(d[v]>d[u]+e[i].dis)
{ if(!action)
{belong[i^1][S]=belong[i][S]=1;}
d[v]=d[u]+e[i].dis;Q.push((heap){v,d[v]});
}
}
}
if(!action)
{
for(int i=1;i<=n;++i)
if(d[i]<oo)sum_all+=d[i];
else sum_all+=L;
}
else
{
for(int i=1;i<=n;++i)
{
if(dis[S][i]<oo)sum_mid-=dis[S][i];
else sum_mid-=L;///去除原边的影响
if(d[i]<oo)sum_mid+=d[i];
else sum_mid+=L;///加上新边
}
}
}
int main()
{ int a,b,c;
while(~scanf("%d%d%d",&n,&m,&L))
{ data();
for(int i=0;i<m;++i)
{
scanf("%d%d%d",&a,&b,&c);
add(a,b,c);add(b,a,c);
}
sum_all=0;
memset(del,0,sizeof(del));
memset(belong,0,sizeof(belong));
for(int i=1;i<=n;++i)
dijstra(dis[i],i,0);
sum_del=sum_all;
for(int i=0;i<m;++i)
{ sum_mid=sum_all;
del[i+i]=del[i+i+1]=1;
for(int j=1;j<=n;++j)
if(belong[i+i][j])///这个优化很关键，有点难，源点固定的最短路会改变，当且仅当最短路上的边
dijstra(dd,j,1);///发生变动，而一条n点最短路最多n-1条边，因此最坏需要n*(n-1)次改变
if(sum_del<sum_mid)sum_del=sum_mid;
del[i+i]=del[i+i+1]=0;
}
printf("%lld %lld\n",sum_all,sum_del);
}
return 0;
}