七、图(中):Saving James Bond

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题目描述

This time let us consider the situation in the movie “Live and Let Die” in which James Bond, the world’s most famous spy, was captured by a group of drug dealers. He was sent to a small piece of land at the center of a lake filled with crocodiles. There he performed the most daring action to escape – he jumped onto the head of the nearest crocodile! Before the animal realized what was happening, James jumped again onto the next big head… Finally he reached the bank before the last crocodile could bite him (actually the stunt man was caught by the big mouth and barely escaped with his extra thick boot).

Assume that the lake is a 100 by 100 square one. Assume that the center of the lake is at (0,0) and the northeast corner at (50,50). The central island is a disk centered at (0,0) with the diameter of 15. A number of crocodiles are in the lake at various positions. Given the coordinates of each crocodile and the distance that James could jump, you must tell him a shortest path to reach one of the banks. The length of a path is the number of jumps that James has to make.

Input Specification:
Each input file contains one test case. Each case starts with a line containing two positive integers N (≤100), the number of crocodiles, and D, the maximum distance that James could jump. Then N lines follow, each containing the (x,y) location of a crocodile. Note that no two crocodiles are staying at the same position.

Output Specification:
For each test case, if James can escape, output in one line the minimum number of jumps he must make. Then starting from the next line, output the position (x,y) of each crocodile on the path, each pair in one line, from the island to the bank. If it is impossible for James to escape that way, simply give him 0 as the number of jumps. If there are many shortest paths, just output the one with the minimum first jump, which is guaranteed to be unique.

Sample Input 1:

Sample Output 1:

Sample Input 2:

Sample Output 2:

代码

#include<stdio.h>
#include<malloc.h>
#include<math.h>
#define MaxVertexNum 100
#define INFINITYMAX 65535
#define Radius 7.5
typedef int Vertex;
typedef double WeightType;
int path[MaxVertexNum][MaxVertexNum];
int MaxDist;
WeightType D[MaxVertexNum][MaxVertexNum];typedef struct ENode *PtrToENode;
struct ENode{Vertex V1, V2;WeightType Weight;
};
typedef PtrToENode Edge;typedef struct GNode *PtrToGNode;
struct GNode{int Nv;int Ne;WeightType G[MaxVertexNum][MaxVertexNum];
};
typedef PtrToGNode MGraph;struct Location{int x;int y;
}Loc[MaxVertexNum];typedef struct MinPath *PtrToMinPath;
struct MinPath{/* 存储最短路径 */Vertex FirstNode;Vertex LastNode;WeightType Dist;
};MGraph CreateGraph( int VertexNum )
{Vertex V, W;MGraph Graph;Graph = (MGraph)malloc(sizeof(struct GNode));Graph->Nv = VertexNum;Graph->Ne = 0;for( V=0; V!=Graph->Nv; ++V )for( W=0; W!= Graph->Nv; ++W )Graph->G[V][W] = INFINITYMAX;return Graph;
}void InsertEdge( MGraph Graph, Edge E )
{Graph->G[E->V1][E->V2] = E->Weight;Graph->G[E->V2][E->V1] = E->Weight;
}double CalDist( struct Location Loc1, struct Location Loc2 )
{return sqrt( pow(Loc1.x-Loc2.x,2) + pow(Loc1.y-Loc2.y,2) );	
}MGraph BuildGraph()
{MGraph Graph;Edge E;int Nv, i, j=0, x, y;scanf("%d %d",&Nv,&MaxDist);for( i=0; i!=Nv; ++i ){scanf("\n%d %d",&x,&y);if ( sqrt(pow(x,2)+pow(y,2))>=Radius && x<=50 && x>=-50 && y<=50 && y>=-50 ){Loc[j].x = x;Loc[j].y = y;++j;}}Graph = CreateGraph( j-1 );E = (Edge)malloc(sizeof(struct ENode));for( i=0; i!=Graph->Nv; ++i )for ( j=i; j!=Graph->Nv; ++j ){if( i!=j ){E->V1 = i;E->V2 = j;if(CalDist(Loc[i], Loc[j]) <= MaxDist)E->Weight = CalDist( Loc[i], Loc[j] );else E->Weight = INFINITYMAX;InsertEdge(Graph, E); }}	return Graph;
}void Floyd( MGraph Graph, WeightType D[][MaxVertexNum] )
{Vertex i, j ,k;for( i=0; i!=Graph->Nv; ++i )for ( j = 0; j!=Graph->Nv; ++j ){D[i][j] = Graph->G[i][j];path[i][j] = -1;}for( k=0; k!=Graph->Nv; ++k )for( i=0; i!=Graph->Nv; ++i )for( j=0; j!=Graph->Nv; ++j )if( D[i][k] + D[k][j] < D[i][j] ){D[i][j] = D[i][k] + D[k][j];path[i][j] = k;}
}int FirstJump(struct Location Loc)
{if(pow(Radius+MaxDist,2)>=pow(Loc.x,2)+pow(Loc.y,2))return 1;else return 0;
}WeightType LastJump(Vertex i)
{int min = abs(50-Loc[i].x);if(abs(Loc[i].y+50)<min)min = abs(Loc[i].y+50);if(abs(50-Loc[i].y)<min)min = abs(50-Loc[i].y);if(abs(50+Loc[i].y)<min)min = abs(50+Loc[i].y);return min;
}PtrToMinPath Compare(Vertex i, Vertex j, PtrToMinPath P)
{WeightType LastDist = LastJump(i);WeightType DD = D[i][j]+sqrt(pow(Loc[j].x,2)+pow(Loc[j].y,2)) + LastDist;if(i==j)DD = sqrt(pow(Loc[j].x,2)+pow(Loc[j].y,2)) + LastDist;if ( (DD < P->Dist) || ( DD == P->Dist && pow(Loc[j].x,2)+pow(Loc[j].y,2) < pow(Loc[P->FirstNode].x,2)+pow(Loc[P->FirstNode].y,2) ) ){/* 如果从i到j的距离小于最小距离，或者等于最小距离同时第一跳的距离最短 */P->FirstNode = j;P->LastNode = i;P->Dist = DD;}return P;
}struct Location Queue[MaxVertexNum];
Vertex last = -1;void ListNode(Vertex i, Vertex j)
{while(1){Queue[++last] = Loc[j];if(i==j)break;j = path[i][j];if(j==-1){Queue[++last] = Loc[i];break;}}printf("%d\n",last+2);while(last!=-1){printf("%d %d\n",Queue[last].x,Queue[last].y);--last;}
}int main()
{Vertex i, j;PtrToMinPath P;P = (PtrToMinPath)malloc(sizeof(struct MinPath));P->Dist = INFINITYMAX;MGraph Graph = BuildGraph();if(MaxDist+Radius>=50){printf("1\n");return 0;}Floyd(Graph, D);for( i=0; i!=Graph->Nv; ++i ){/* 扫描所有结点，找到能跳到岸上的结点 */if(Loc[i].x>=50-MaxDist || Loc[i].x<=MaxDist-50 || Loc[i].y>=50-MaxDist || Loc[i].y<=MaxDist-50){for(j=0; j!=Graph->Nv; ++j)if(FirstJump(Loc[j]))/* 扫描所有结点，找到第一步能跳到的结点，并判断跳到岸上的距离是不是最小 */P = Compare(i,j,P);}}if(P->Dist == INFINITYMAX)printf("0\n");else ListNode(P->FirstNode,P->LastNode);return 0;
}

出现的问题和解决方法

题目看错了，人家问的是跳跃次数最少的路径，我写成了跳跃距离最短的路径，神奇的是只有一个测试点“最大N，sample1的复杂版，可选路径8条，后面要更新前面的最短路”通不过。改了一下午，今天天气特别热，实在是不行了，懒得改了，就按最小距离得了。
另外，如果鳄鱼在小岛上，詹姆斯邦德不是直接就被吃了吗，所以这种情况按理说应该直接输出0就对了啊。有问题。

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