编程练习7（最短路）

A:Frogger(POJ2253)

Description

Freddy Frog is sitting on a stone in the middle of a lake. Suddenly he notices Fiona Frog who is sitting on another stone. He plans to visit her, but since the water is dirty and full of tourists’ sunscreen, he wants to avoid swimming and instead reach her by jumping.
Unfortunately Fiona’s stone is out of his jump range. Therefore Freddy considers to use other stones as intermediate stops and reach her by a sequence of several small jumps.
To execute a given sequence of jumps, a frog’s jump range obviously must be at least as long as the longest jump occuring in the sequence.
The frog distance (humans also call it minimax distance) between two stones therefore is defined as the minimum necessary jump range over all possible paths between the two stones.

You are given the coordinates of Freddy’s stone, Fiona’s stone and all other stones in the lake. Your job is to compute the frog distance between Freddy’s and Fiona’s stone.

Input

The input will contain one or more test cases. The first line of each test case will contain the number of stones n (2<=n<=200). The next n lines each contain two integers xi,yi (0 <= xi,yi <= 1000) representing the coordinates of stone #i. Stone #1 is Freddy’s stone, stone #2 is Fiona’s stone, the other n-2 stones are unoccupied. There’s a blank line following each test case. Input is terminated by a value of zero (0) for n.

Output

For each test case, print a line saying “Scenario #x” and a line saying “Frog Distance = y” where x is replaced by the test case number (they are numbered from 1) and y is replaced by the appropriate real number, printed to three decimals. Put a blank line after each test case, even after the last one.

Sample Input

2
0 0
3 4

3
17 4
19 4
18 5

0

Sample Output

Scenario #1
Frog Distance = 5.000

Scenario #2
Frog Distance = 1.414

我的代码

#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#include<string>#include<vector>#include<stack>#include<bitset>#include<cstdlib>#include<cmath>#include<set>#include<list>#include<deque>#include<map>#include<queue>using namespace std;typedef long long ll;int T,num,n,m;int sal,p;int i,j,k;struct Node{    int x,y;};struct edge{    int u,v;    double w;};const double INF = 1e8;Node stone[205];bool visited[205];double d[205];double distance(int a, int b){    return sqrt(double((stone[a].x-stone[b].x)*(stone[a].x-stone[b].x)+(stone[a].y-stone[b].y)*(stone[a].y-stone[b].y)));}void init(){    for (i = 1 ; i <= num; i++) {        d[i] = INF;        visited[i] = false;    }    d[1] = 0;    visited[1] = true;}void dijstra(){    int v = 1;    init();    for (i = 1 ; i <= num; i++) {        int min = INF;        //找到最小的d        for (j = 1; j <= num; j++) {            if(!visited[j] && d[j] < min){                min = d[j];                v = j;            }        }       //标记为访问过，更新其他点        visited[v] = true;        if (v == 2) {            printf("Scenario #%d\n",T);            printf("Frog Distance = %.3f\n\n",d[2]);            return;        }        for (j = 1; j <= num ; j++) {            if (!visited[j] &&  d[j] > max(d[v],distance(v,j))) {                d[j] = max(d[v],distance(v,j));            }        }    }}int main() {   // freopen("input.txt", "r", stdin);    T = 0;    while(scanf("%d",&num)!= EOF && num != 0) {        T++;        for (i = 1; i <= num; i++) {            scanf("%d%d",&stone[i].x,&stone[i].y);        }        dijstra();    }    return 0;}

B:Wormholes(POJ3259)

Description

While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it is a one-way path that delivers you to its destination at a time that is BEFORE you entered the wormhole! Each of FJ’s farms comprises N (1 ≤ N ≤ 500) fields conveniently numbered 1..N, M (1 ≤ M ≤ 2500) paths, and W (1 ≤ W ≤ 200) wormholes.

As FJ is an avid time-traveling fan, he wants to do the following: start at some field, travel through some paths and wormholes, and return to the starting field a time before his initial departure. Perhaps he will be able to meet himself :) .

To help FJ find out whether this is possible or not, he will supply you with complete maps to F (1 ≤ F ≤ 5) of his farms. No paths will take longer than 10,000 seconds to travel and no wormhole can bring FJ back in time by more than 10,000 seconds.

Input

Line 1: A single integer, F. F farm descriptions follow.
Line 1 of each farm: Three space-separated integers respectively: N, M, and W
Lines 2.. M+1 of each farm: Three space-separated numbers ( S, E, T) that describe, respectively: a bidirectional path between S and E that requires T seconds to traverse. Two fields might be connected by more than one path.
Lines M+2.. M+ W+1 of each farm: Three space-separated numbers ( S, E, T) that describe, respectively: A one way path from S to E that also moves the traveler back T seconds.

Output

Lines 1.. F: For each farm, output “YES” if FJ can achieve his goal, otherwise output “NO” (do not include the quotes).

Sample Input

2
3 3 1
1 2 2
1 3 4
2 3 1
3 1 3
3 2 1
1 2 3
2 3 4
3 1 8

Sample Output

NO
YES

我的代码

#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#include<string>#include<vector>#include<stack>#include<bitset>#include<cstdlib>#include<cmath>#include<set>#include<list>#include<deque>#include<map>#include<queue>using namespace std;typedef long long ll;const int INF = 1000000000;int numf,n,m;int N, M, W;int i,j;struct Edge{    int beg,end;    int w;};int d[505];Edge edge[7000];//w表示边数void init(){    for (i = 1; i <=N; i++) {        d[i] = INF;    }    d[1] = 0;}bool bellman(int w){    for (i = 1; i <= N; i++) {        for (j = 1; j <= w; j++) {            if (d[edge[j].end] > d[edge[j].beg] + edge[j].w ) {                d[edge[j].end] = d[edge[j].beg] + edge[j].w;            }        }    }    for (j = 1; j <= w; j++) {        if (d[edge[j].end] > d[edge[j].beg] + edge[j].w) {            return false;        }    }    return true;}int main() {    //freopen("input.txt", "r", stdin);    scanf("%d",&numf);    while(numf--) {        init();        scanf("%d%d%d",&N,&M,&W);        int k = 1;        for (i = 1; i <= M; i++) {            scanf("%d%d%d",&edge[k].beg,&edge[k].end,&edge[k].w);            k++;            edge[k].beg = edge[k-1].end;edge[k].end = edge[k-1].beg;edge[k].w = edge[k-1].w;            k++;        }        for (i = 1; i <= W; i++) {            scanf("%d%d%d",&edge[k].beg,&edge[k].end,&edge[k].w);            edge[k].w = - edge[k].w;            k++;        }        bool flag = bellman(2*M+W);        if (!flag) {            printf("YES\n");        }else{            printf("NO\n");        }    }    return 0;}