1 /* 2 ?*題目大意： 3 ?*在一個(gè)有向圖中,求從s到t兩個(gè)點(diǎn)之間的最短路和比最短路長1的次短路的條數(shù)之和; 4 ?* 5 ?*算法思想： 6 ?*用A*求第K短路,目測會超時(shí),直接在dijkstra算法上求次短路; 7 ?*將dist數(shù)組開成二維的,即dist[v][2],第二維分別用于記錄最短路和次短路; 8 ?*再用一個(gè)cnt二維數(shù)組分別記錄最短路和次短路的條數(shù); 9 ?*每次更新路徑的條數(shù)時(shí),不能直接加1,,應(yīng)該加上cnt[u][k],k為次短路徑或者最短路徑的標(biāo)記; 10 ?*圖有重邊,不能用鄰接矩陣存儲; 11 ?*不知道為什么,題目上說的是N and M, separated by a single space, with 2≤N≤1000 and 1 ≤ M ≤ 10000; 12 ?*而我的代碼硬是把N開成1W了才過,求解釋,RE了無數(shù)次,擦; 13 **/? 14 #include<iostream>? 15 #include<cstdio>? 16 #include<cstring>? 17 #include<string>? 18 #include<algorithm>? 19 using namespace std;? 20 ??? 21 const int N=11111;? 22 const int M=111111;? 23 const int INF=0xffffff;? 24 ??? 25 struct node? 26 {? 27 ????int to;? 28 ????int w;? 29 ????int next;? 30 };? 31 ??? 32 node edge[N];? 33 int head[N];? 34 ??? 35 int dist[N][2],cnt[N][2];? 36 bool vis[N][2];? 37 int n,m,s,t,edges;? 38 ??? 39 void addedge(int u,int v,int w)? 40 {? 41 ????edge[edges].w=w;? 42 ????edge[edges].to=v;? 43 ????edge[edges].next=head[u];? 44 ????head[u]=edges++;? 45 }? 46 ??? 47 int dijkstra()? 48 {? 49 ????int k;? 50 ????for(int i=0; i<=n; i++)? 51 ????{? 52 ????????dist[i][0]=dist[i][1]=INF;? 53 ????????vis[i][0]=vis[i][1]=0;? 54 ????????cnt[i][0]=cnt[i][1]=0;? 55 ????}? 56 ????cnt[s][0]=1,dist[s][0]=0;? 57 ??? 58 ????for(int i=1; i<=n*2; i++)? 59 ????{? 60 ????????int u=-1;? 61 ????????int min_dist=INF;? 62 ????????for(int j=1; j<=n; j++)? 63 ????????????for(int flag=0; flag<2; flag++)? 64 ????????????????if(!vis[j][flag]&&dist[j][flag]<min_dist)? 65 ????????????????{? 66 ????????????????????min_dist=dist[j][flag];? 67 ????????????????????u=j;? 68 ????????????????????k=flag;? 69 ????????????????}? 70 ????????if(u==-1)? 71 ????????????break;? 72 ????????vis[u][k]=true;? 73 ????????for(int e=head[u]; e!=-1; e=edge[e].next)? 74 ????????{? 75 ????????????int j=edge[e].to;? 76 ????????????int tmp=dist[u][k]+edge[e].w;? 77 ??? 78 ????????????if(tmp<dist[j][0])//tmp小于最短路徑長:? 79 ????????????{? 80 ????????????????dist[j][1]=dist[j][0];//次短路徑長? 81 ????????????????cnt[j][1]=cnt[j][0];//次短路徑計(jì)數(shù)? 82 ????????????????dist[j][0]=tmp;//最短路徑長? 83 ????????????????cnt[j][0]=cnt[u][k];//最短路徑計(jì)數(shù)? 84 ????????????}? 85 ??? 86 ????????????else if(tmp==dist[j][0])//tmp等于最短路徑長：? 87 ????????????{? 88 ????????????????cnt[j][0]+=cnt[u][k];//最短路徑計(jì)數(shù)? 89 ????????????}? 90 ??? 91 ????????????else if(tmp<dist[j][1])//tmp大于最短路徑長且小于次短路徑長：? 92 ????????????{? 93 ????????????????dist[j][1]=tmp;//次短路徑長? 94 ????????????????cnt[j][1]=cnt[u][k];//次短路徑計(jì)數(shù)? 95 ????????????}? 96 ??? 97 ????????????else if(tmp==dist[j][1])//tmp等于次短路徑長：? 98 ????????????{? 99 ????????????????cnt[j][1]+=cnt[u][k];//次短路徑計(jì)數(shù)? 100 ????????????}? 101 ????????}? 102 ????}? 103 ??? 104 ????int res=cnt[t][0];? 105 ????if(dist[t][0]+1==dist[t][1])//判斷最短路和次短路是否相差1? 106 ????????res+=cnt[t][1];? 107 ????return res;? 108 }? 109 ??? 110 int main()? 111 {? 112 ????//freopen("C:\\Users\\Administrator\\Desktop\\kd.txt","r",stdin);? 113 ????int tcase;? 114 ????scanf("%d",&tcase);? 115 ????while(tcase--)? 116 ????{? 117 ????????edges=0;? 118 ????????scanf("%d%d",&n,&m);? 119 ????????memset(head,-1,sizeof(head));? 120 ????????int u,v,w;? 121 ????????for(int i=0; i<m; i++)? 122 ????????{? 123 ????????????scanf("%d%d%d",&u,&v,&w);? 124 ????????????addedge(u,v,w);? 125 ????????}? 126 ????????scanf("%d%d",&s,&t);? 127 ????????printf("%d\n",dijkstra());? 128 ????}? 129 ????return 0;? 130 }?

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轉(zhuǎn)載于:https://www.cnblogs.com/Hammer-cwz-77/p/7337212.html

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