Day4-A-最短路 HDU2544
在每年的校赛里,所有进入决赛的同学都会获得一件很漂亮的t-shirt。但是每当我们的工作人员把上百件的衣服从商店运回到赛场的时候,却是非常累的!所以现在他们想要寻找最短的从商店到赛场的路线,你可以帮助他们吗?
Input输入包括多组数据。每组数据第一行是两个整数N、M(N<=100,M<=10000),N表示成都的大街上有几个路口,标号为1的路口是商店所在地,标号为N的路口是赛场所在地,M则表示在成都有几条路。N=M=0表示输入结束。接下来M行,每行包括3个整数A,B,C(1<=A,B<=N,1<=C<=1000),表示在路口A与路口B之间有一条路,我们的工作人员需要C分钟的时间走过这条路。
输入保证至少存在1条商店到赛场的路线。
Output对于每组输入,输出一行,表示工作人员从商店走到赛场的最短时间Sample Input
2 1 1 2 3 3 3 1 2 5 2 3 5 3 1 2 0 0
Sample Output
3 2
简单的最短路板子题,用来练习各种算法,代码如下:
dijkstra:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
const int maxm = 110; const int INF = 0x7fffffff; int N, M, G[maxm][maxm], d[maxm], vis[maxm]; struct Node { int sum, i; Node(int _sum, int _i) : sum(_sum), i(_i){} bool operator<(const Node &a) const { return a.sum < sum; } }; void init() { for (int i = 2; i <= N; ++i) { d[i] = INF; } memset(G, 0, sizeof(G)), memset(vis, 0, sizeof(vis)); } int main() { while(scanf("%d%d",&N,&M) && N + M) { init(); for (int i = 0; i < M; ++i) { int t1, t2, t3; scanf("%d%d%d", &t1, &t2, &t3); G[t1][t2] = G[t2][t1] = t3; } priority_queue<Node>q; q.push(Node(0, 1)); while(!q.empty()) { Node p = q.top(); q.pop(); if(vis[p.i]++) continue; for (int i = 1; i <= N; ++i) { if(G[p.i][i] && G[p.i][i] + d[p.i] < d[i]) { d[i] = G[p.i][i] + d[p.i]; q.push(Node(d[i],i)); } } } printf("%d\n", d[N]); } return 0; }
Bellman_Ford:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
const int maxm = 110; const int maxn = 20010; const int INF = 0x7fffffff; int N, M, v[maxn], u[maxn], cost[maxn], d[maxm]; void init() { memset(v, 0, sizeof(v)), memset(u, 0, sizeof(u)), memset(cost, 0, sizeof(cost)); for (int i = 2; i <= N; ++i) d[i] = INF; } int main() { while(scanf("%d%d",&N,&M) && N + M) { init(); for (int i = 0; i < M; ++i) { int t1, t2, t3; scanf("%d%d%d", &t1, &t2, &t3); v[i * 2] = t1, u[i * 2] = t2, cost[i * 2] = t3; v[i * 2 + 1] = t2, u[i * 2 + 1] = t1, cost[i * 2 + 1] = t3; } for (int j = 0; j < N-1; ++j) { for (int i = 0; i < M * 2; ++i) { int x = v[i], y = u[i], c = cost[i]; if(d[x] < INF) d[y] = min(d[y], d[x] + c); } } printf("%d\n", d[N]); } return 0; }
SPFA:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
const int maxm = 110; const int INF = 0x7fffffff; int N, M, G[maxm][maxm], d[maxm], inq[maxm]; struct Node { int sum, i; Node(int _sum, int _i) : sum(_sum), i(_i){} }; void init() { for (int i = 2; i <= N; ++i) { d[i] = INF; } memset(G, 0, sizeof(G)), memset(inq, 0, sizeof(inq)); } int main() { while(scanf("%d%d",&N,&M) && N + M) { init(); for (int i = 0; i < M; ++i) { int t1, t2, t3; scanf("%d%d%d", &t1, &t2, &t3); G[t1][t2] = G[t2][t1] = t3; } queue<int>q; q.push(1); inq[1] = 1; while(!q.empty()) { int u = q.front(); q.pop(); inq[u] = 0; for (int i = 1; i <= N; ++i) { if(G[u][i] && G[u][i] + d[u] < d[i]) { d[i] = G[u][i] + d[u]; if(!inq[i]) q.push(i); } } } printf("%d\n", d[N]); } return 0; }
Floyd:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
const int maxm = 110; const int INF = 0x7fffffff; int N, M, G[maxm][maxm]; void init() { for(int i = 1; i <= N; ++i) { for (int j = 1; j <= N; ++j) { if(i == j) G[i][i] = 0; else G[i][j] = INF; } } } int main() { while(scanf("%d%d",&N,&M) && N + M) { init(); for (int i = 0; i < M; ++i) { int t1, t2, t3; scanf("%d%d%d", &t1, &t2, &t3); G[t1][t2] = G[t2][t1] = t3; } for (int k = 1; k <= N; ++k) for(int i = 1; i <= N; ++i) for (int j = 1; j <= N; ++j) { if(G[i][k] < INF && G[k][j] < INF) G[i][j] = min(G[i][j], G[i][k] + G[k][j]); } printf("%d\n", G[1][N]); } return 0; }