HDU-2544 最短路【最短路】

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2544

最近复习了最短路径的算法，就写了4个版本的测试。正好是模板题，就果断A之。。。

Dijkstar版本：

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<climits>#include<queue>#include<algorithm>using namespace std;#define N 110#define MAX 999999#define CLR(arr, what) memset(arr, what, sizeof(arr))int nodenum, edgenum;int map[N][N], dis[N];bool visit[N];int Dijkstra(int src, int des){    int temp, k;    CLR(visit, false);    for(int i = 1; i <= nodenum; ++i)        dis[i] = (i == src ? 0 : map[src][i]);    visit[src] = true;    dis[src] = 0;    for(int i = 1; i<= nodenum; ++i)    {        temp = MAX;        for(int j = 1; j <= nodenum; ++j)            if(!visit[j] && temp > dis[j])                temp = dis[k = j];        if(temp == MAX)            break;        visit[k] = true;        for(int j = 1; j <= nodenum; ++j)            if(!visit[j] && dis[j] > dis[k] + map[k][j])                dis[j] = dis[k] + map[k][j];    }    return dis[des];}int main(){    int start, end, cost;    int answer;    while(~scanf("%d%d", &nodenum, &edgenum) && (nodenum + edgenum))    {        for(int i = 1; i <= nodenum; ++i)            for(int j = 1; j <= nodenum; ++j)            map[i][j] = MAX;        for(int i = 1; i <= edgenum; ++i)        {            scanf("%d%d%d", &start, &end, &cost);            if(cost < map[start][end])                map[start][end] = map[end][start] = cost;        }        answer = Dijkstra(1, nodenum);        printf("%d\n", answer);    }    return 0;}

Bellman_Ford版本：

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<climits>#include<queue>#include<algorithm>using namespace std;#define N 110#define MAX 999999#define CLR(arr, what) memset(arr, what, sizeof(arr))int nodenum, edgenum;int map[N][N], dis[N];bool visit[N];struct Edge{    int u, v;    int cost;}e[N * N / 2];int Bellman_ford(int src, int des){    for(int i = 1; i <= nodenum; ++i)        dis[i] = MAX;    dis[src] = 0;    for(int i = 0; i < nodenum - 1; ++i) //n-1遍        for(int j = 0; j < edgenum * 2; ++j) //each edge            if(dis[e[j].v] > dis[e[j].u] + e[j].cost)                dis[e[j].v] = dis[e[j].u] + e[j].cost;    return dis[des];}int main(){    int start, end, cost;    int answer;    while(~scanf("%d%d", &nodenum, &edgenum) && (nodenum + edgenum))    {        for(int i = 0; i < edgenum; ++i)        {            scanf("%d%d%d", &start, &end, &cost); //双向边            e[i * 2].u = start, e[i * 2].v = end, e[i * 2].cost = cost;            e[i * 2 + 1].u = end, e[i * 2 + 1].v = start, e[i * 2 + 1].cost = cost;        }        answer = Bellman_ford(1, nodenum);        printf("%d\n", answer);    }    return 0;}

Floyd版本：

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<climits>#include<queue>#include<algorithm>using namespace std;#define N 110#define MAX INT_MAX >> 1#define CLR(arr, what) memset(arr, what, sizeof(arr))int nodenum, edgenum;int map[N][N], dis[N];bool visit[N];int Floyd(int src, int des) //多源多汇最短路{    for(int k = 1; k <= nodenum; ++k)        for(int i = 1; i <= nodenum; ++i)            for(int j = 1; j <= nodenum; ++j)                map[i][j] = min(map[i][j], map[i][k] + map[k][j]);    return map[src][des];}int main(){    int start, end, cost;    int answer;    while(~scanf("%d%d", &nodenum, &edgenum) && (nodenum + edgenum))    {        for(int i = 1; i <= nodenum; ++i)            for(int j = 1; j <= nodenum; ++j)                map[i][j] = MAX;        for(int i = 0; i < edgenum; ++i)        {            scanf("%d%d%d", &start, &end, &cost);            if(cost < map[start][end])                map[start][end] = map[end][start] = cost;        }        answer = Floyd(1, nodenum);        printf("%d\n", answer);    }    return 0;}

SPFA版本：

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<climits>#include<queue>#include<algorithm>using namespace std;#define N 110#define MAX INT_MAX >> 1#define CLR(arr, what) memset(arr, what, sizeof(arr))int nodenum, edgenum;int map[N][N], dis[N];bool visit[N];int SPFA(int src, int des){    queue<int> q;    CLR(visit, false);    for(int i = 1;i <= nodenum; ++i)        dis[i] = MAX;    dis[src] = 0;    visit[src] = true;    q.push(src);    while(!q.empty())    {        int cur = q.front();        q.pop();        visit[cur] = false; //出队标记为false        for(int i = 1; i <= nodenum; ++i)        {            if(dis[i] > dis[cur] + map[cur][i]) //没有2个集合，和Dijkstra有本质区别            {                dis[i] = dis[cur] + map[cur][i]; //能松弛就松弛                if(!visit[i]) //不在队列中则加入，然后更新所有以前经过此点的最短路径                {                    q.push(i);                    visit[i] = true;                }            }        }    }    return dis[des];}int main(){    int start, end, cost;    int answer;    while(~scanf("%d%d", &nodenum, &edgenum) && (nodenum + edgenum))    {        for(int i = 1; i <= nodenum; ++i)            for(int j = 1; j <= nodenum; ++j)                map[i][j] = MAX;        for(int i = 0; i < edgenum; ++i)        {            scanf("%d%d%d", &start, &end, &cost);            if(cost < map[start][end])                map[start][end] = map[end][start] = cost;        }        answer = SPFA(1, nodenum);        printf("%d\n", answer);    }    return 0;}