題目鏈接

http://codeforces.com/gym/101630/attachments

題解

zyb學(xué)長(zhǎng)的題。
先枚舉第\(k\)大的邊權(quán)，設(shè)其邊權(quán)為\(x\)，然后把每條邊邊權(quán)減掉\(x\)與\(0\)取\(\max\), 跑最短路之后加上\(x\times k\)更新答案。
注意從\(0\)開(kāi)始枚舉（就相當(dāng)于裸跑最短路）。

代碼

#include<cstdio> #include<cstdlib> #include<iostream> #include<cassert> #include<algorithm> #include<queue> #include<cstring> #define llong long long using namespace std;const int N = 3000; const int M = 6000; struct Edge {int v,nxt; llong w; } e[M+3]; int fe[N+3]; struct AEdge {int u,v; llong w; } ae[M+3]; struct DijNode {int u; llong dis;DijNode() {}DijNode(int _u,llong _dis) {u = _u,dis = _dis;}bool operator <(const DijNode &arg) const {return dis>arg.dis;} }; priority_queue<DijNode> pq; llong dis[N+3]; bool vis[N+3]; int n,en,p,s,t,m;void addedge(int u,int v,llong w) {en++; e[en].v = v; e[en].w = w;e[en].nxt = fe[u]; fe[u] = en; }void clear() {for(int i=1; i<=m; i++) e[i].v = e[i].nxt = e[i].w = 0;for(int i=1; i<=n; i++) fe[i] = 0;en = 0; }llong Dijkstra() {memset(dis,42,sizeof(dis)); memset(vis,false,sizeof(vis));dis[s] = 0ll; pq.push(DijNode(s,0));while(!pq.empty()){DijNode tmp = pq.top(); pq.pop(); int u = tmp.u;if(tmp.dis!=dis[u]) continue;if(vis[u]==true) continue;vis[u] = true;for(int i=fe[u]; i; i=e[i].nxt){int v = e[i].v;if(dis[v]>dis[u]+e[i].w && vis[v]==false){dis[v] = dis[u]+e[i].w;pq.push(DijNode(v,dis[v]));}}}return dis[t]; }int main() {scanf("%d%d%d",&n,&m,&p); s = 1; t = n;for(int i=1; i<=m; i++){scanf("%d%d%I64d",&ae[i].u,&ae[i].v,&ae[i].w);}llong ans = 10000000000000000ll;for(int i=0; i<=m; i++){for(int j=1; j<=m; j++) addedge(ae[j].u,ae[j].v,max(ae[j].w-ae[i].w,0ll)),addedge(ae[j].v,ae[j].u,max(ae[j].w-ae[i].w,0ll));llong cur = Dijkstra();ans = min(ans,cur+ae[i].w*p);clear();}printf("%I64d\n",ans);return 0; }

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