題意：

　　n個點,m條固定邊，k條可選邊，可選邊最多只能選擇一條，也可以不選擇，問起點s到終點t的最小權值是什麽？

題解：

1.正確的解法：

　　以起點終點為源點，求固定邊最短路，然後對於k條可選邊，比如可選邊為起點p,終點q，則含有本條可選邊的最優路徑的距離為ds[p]+dt[q]+w(w為可選邊權值，ds[p]為p到起點最短距離，dt[q]為q到終點最短距離)

　　復雜度:mlogn+k

2.歪解：

　　自己寫看到這道題n只有500，m只有1000，直接求K次最短路，暴力求解，之後過了。。。。。。

坑點：

　　每組數據輸出時中間隔一行，最後一組數據之後不要空行

代碼：

#include<iostream>
using
 
 namespace std;
#include<cstdio>
#include<cstdlib>
#include<vector>
#define MAXN 1000
#include<vector>
#include<queue>
typedef long long LL;
struct teamdata{
    LL d;LL point;
    bool operator<(const teamdata& pt)const{
        return this->d>pt.d;
    }
};
 
struct bian{
    LL d,point;
};
LL n,m,k,l,s,t;
LL pre[MAXN];
vector<LL> ansl;
vector<bian> maps[MAXN];
const LL inf=0x3f3f3f3f;
LL d[MAXN];
queue<teamdata> team;
LL SPFA(LL s,LL t){
    LL ans=-1;
    teamdata pt;pt.d=0;pt.point=s;
    while(!team.empty()) team.pop();
    team.push(pt);teamdata pd;
    
 
for (LL i=0;i<=MAXN-10;i++) d[i]=inf,pre[i]=-1;d[s]=0;
    while(!team.empty()){
        pt=team.front();team.pop();
        if (pt.d>d[pt.point]) continue;
        if (pt.point==t){
            ans=pt.d;continue; 
        } 
        for (LL i=0;i<maps[pt.point].size();i++){
            LL nb=maps[pt.point][i].point;
            if (d[nb]>pt.d+maps[pt.point][i].d){
                d[nb]=pt.d+maps[pt.point][i].d;
                pre[nb]=pt.point;
                pd.d=d[nb];pd.point=nb;team.push(pd);
            }
        }
    }
    return ans;
}
int main(){
    int sp=0;
    while(scanf("%lld%lld%lld",&n,&s,&t)!=EOF){
        sp++;
        if (sp>1) putchar(‘\n‘);
        scanf("%lld",&m);
        bian pt;
        for (int i=0;i<=n;i++) maps[i].clear();
        for (LL i=0;i<m;i++){
            LL p,q,k;
            scanf("%lld%lld%lld",&p,&q,&k);
            pt.d=k;pt.point=q;
            maps[p].push_back(pt);
            pt.point=p;
            maps[q].push_back(pt);
        }
        scanf("%lld",&k);
        LL zans=SPFA(s,t);
        ansl.clear();
                LL x=t;
                while(x>0) {
                    ansl.push_back(x);
                    x=pre[x];
                }
        LL bj=-1,bjm=-1;
        for (LL i=0;i<k;i++){
            LL p,q,k;
            scanf("%lld%lld%lld",&p,&q,&k);
            bian pt;pt.d=k;pt.point=q;
            maps[p].push_back(pt);
            pt.point=p;
            maps[q].push_back(pt);
            LL zz=SPFA(s,t);
            if (zz<zans&&zz>=0) {
                zans=zz;
                if (pre[q]==p) bj=p,bjm=q;
                else bj=q,bjm=p;
                ansl.clear();
                LL x=t;
                while(x>0) {
                    ansl.push_back(x);
                    x=pre[x];
                }
            }
            maps[p].pop_back();
            maps[q].pop_back();
        }
        int ppt=-1,used=0;
        for (LL i=ansl.size()-1;i>=0;i--){
            printf("%lld",ansl[i]);
            if (i>0) putchar(‘ ‘);
            if (bjm==ansl[i]&&bj==ppt) used=1;
            ppt=ansl[i];
        }
        putchar(‘\n‘);
        if (used){
        printf("%lld\n",bj);
        }
        else{
            printf("Ticket Not Used\n");
        }
        printf("%lld\n",zans);
    }
    return 0;
}

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