http://poj.org/problem?id=1511
题意:(图论147)。求从起点到各个点的最短回路。
分析:用两次spfa就行了。第一次spfa是从出发点start到各个点,求一次最短距离,第二次spfa是从各个点反向求到出发点的最短距离。
处理:第一次spfa只要保存正向边,第二次用反向边进行spfa就可以了。最后把两次求得的距离相加就是最短回路。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
// I'm the Topcoder //C #include <stdio.h> #include <stdlib.h> #include <string.h> #include <ctype.h> #include <math.h> #include <time.h> //C++ #include <iostream> #include <algorithm> #include <cstdio> #include <cstdlib> #include <cmath> #include <cstring> #include <cctype> #include <stack> #include <string> #include <list> #include <queue> #include <map> #include <vector> #include <deque> #include <set> using namespace std; //*************************OUTPUT************************* #ifdef WIN32 #define INT64 "%I64d" #define UINT64 "%I64u" #else #define INT64 "%lld" #define UINT64 "%llu" #endif //**************************CONSTANT*********************** #define INF 0x3f3f3f3f #define eps 1e-8 #define PI acos(-1.) #define PI2 asin (1.); typedef long long LL; //typedef __int64 LL; //codeforces typedef unsigned int ui; typedef unsigned long long ui64; #define MP make_pair typedef vector<int> VI; typedef pair<int, int> PII; #define pb push_back #define mp make_pair //***************************SENTENCE************************ #define CL(a,b) memset (a, b, sizeof (a)) #define sqr(a,b) sqrt ((double)(a)*(a) + (double)(b)*(b)) #define sqr3(a,b,c) sqrt((double)(a)*(a) + (double)(b)*(b) + (double)(c)*(c)) //****************************FUNCTION************************ template <typename T> double DIS(T va, T vb) { return sqr(va.x - vb.x, va.y - vb.y); } template <class T> inline T INTEGER_LEN(T v) { int len = 1; while (v /= 10) ++len; return len; } template <typename T> inline T square(T va, T vb) { return va * va + vb * vb; } // aply for the memory of the stack //#pragma comment (linker, "/STACK:1024000000,1024000000") //end const int maxn = 1000000+100; struct node{ int to; int next; long long weight; }; node edge[maxn],edge1[maxn];//保存边的起点和终点 int n,m; long long val; int tot,tot1; int src;//起点 int head[maxn],head1[maxn]; int visit[maxn],visit1[maxn]; long long dis[maxn],dis1[maxn]; void add(int a,int b,long long c){ edge[tot].to=b; edge[tot].weight=c; edge[tot].next=head[a]; head[a]=tot++; } void add1(int a,int b,long long c){ edge1[tot1].to=b; edge1[tot1].weight=c; edge1[tot1].next=head1[a]; head1[a]=tot1++; } void spfa(){ //初始化 for(int i=1;i<=n;i++){ dis[i]=INF; visit[i]=0;//访问标记 } dis[src]=0; visit[src]=1; int u; int v; queue<int> Q;//优先队列 Q.push(src); while(!Q.empty()){ u=Q.front(); Q.pop(); visit[u]=0;//必须是0,这题是1也能过不过是错的 // visit[u]=1; for(int i=head[u];i!=-1;i=edge[i].next){ v=edge[i].to; if(dis[v]>dis[u]+edge[i].weight){ dis[v]=dis[u]+edge[i].weight; if(!visit[v]){ Q.push(v); visit[v]=1; } } } } } void spfa1(){ //初始化 for(int i=1;i<=n;i++){ dis1[i]=INF; visit1[i]=0; } dis1[src]=0; visit1[src]=1; int u,v; queue<int> Q; Q.push(src); while(!Q.empty()){ u=Q.front(); Q.pop(); visit[u]=0;//必须是0,这题是1也能过不过是错的 //visit1[u]=1; for(int i=head1[u];i!=-1;i=edge1[i].next){ v=edge1[i].to; if(dis1[v]>dis1[u]+edge1[i].weight){ dis1[v]=dis1[u]+edge1[i].weight; if(!visit1[v]){ Q.push(v); visit1[v]=1; } } } } } int main(){ int a,b,cas; scanf("%d",&cas); while(cas--){ scanf("%d%d",&n,&m); tot=tot1=0;//边的条数 for(int i=1;i<=n;i++){ head[i]=-1; head1[i]=-1; } for(int i=1;i<=m;i++){ scanf("%d%d%lld",&a,&b,&val); add(a,b,val);//正向边 add1(b,a,val);//反向边 } src=1;//起点(终点) spfa(); spfa1(); long long sum=0; for(int i=2;i<=n;i++){ sum+=dis[i]+dis1[i]; } printf("%lld\n",sum); } return 0; }