UVALive - 6885 Flowery Trails 最短路

题目链接：

http://acm.hust.edu.cn/vjudge/problem/129723

Flowery Trails

Time Limit: 3000MS

题意

求所有位于最短路上的边的权值和的两倍。

题解

起点跑一遍单源最短路，得到距离数组d[],终点跑一遍最短路，得到d2[]。枚举每条边，如果d[u]+d2[v]+wd[n-1]||d[v]+d2[u]+wd[n-1]，那么ans+=w;

代码

include

include

include

include

include

include

include

include

include

using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) ;//cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)

typedef long long LL;
typedef vector VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-8;

//start----------------------------------------------------------------------

const int maxn=10000+10;

struct Edge{
int u,v,d;
Edge(int u,int v,int d):u(u),v(v),d(d){}
};

struct HeapNode{
int d,u;
HeapNode(int d,int u):d(d),u(u){}
bool operator < (const HeapNode&tmp) const {
return d>tmp.d;
}
};

struct Dijkstra{
int n,m;
vector egs;
vector G[maxn];
bool done[maxn];
int d[maxn],d2[maxn];
vector p[maxn];

void init(int n){
this->n=n;
for(int i=0;i<=n;i++) G[i].clear(),p[i].clear();
egs.clear();
}

void addEdge(int u,int v,int d){
egs.push_back(Edge(u,v,d));
m=egs.size();
G[u].push_back(m-1);
}

void dijkstra(int *d,int s){
priority_queue Q;
rep(i,0,n) d[i]=INF;
d[s]=0;
clr(done,0);
Q.push(HeapNode(0,s));
while(!Q.empty()){
HeapNode x=Q.top(); Q.pop();
int u=x.u;
if(done[u]) continue;
done[u]=1;
rep(i,0,G[u].size()){
Edge& e=egs[G[u][i]];
if(d[e.v]>d[u]+e.d){
d[e.v]=d[u]+e.d;
Q.push(HeapNode(d[e.v],e.v));
}
}
}
}

void solve(){
dijkstra(d,0);
dijkstra(d2,n-1);
LL ans=0;
rep(i,0,m){
Edge& e=egs[i];
if(d[e.u]+d2[e.v]+e.dd[n-1]||d[e.v]+d2[e.u]+e.dd[n-1]){
ans+=e.d;
}
}
printf("%lld\n",ans);
}
}dij;

int n,m;

int main() {
while(scanf("%d%d",&n,&m)==2){
dij.init(n);
rep(i,0,m){
int u,v,w;
scanf("%d%d%d",&u,&v,&w);
dij.addEdge(u,v,w);
dij.addEdge(v,u,w);
}
dij.solve();
}
return 0;
}

//end-----------------------------------------------------------------------

Notes

起点跑一次，终点跑一次，然后枚举边来判断最短路上的边的方法不错。