POJ 1511 spfa求最短回路

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。

View Code

// 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;   //codeforcestypedef unsigned int ui;typedef unsigned long long ui64;#define MP make_pairtypedef 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")//endconst 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;}