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题目链接:.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 edgeif(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falsefor(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;
}




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