PTA 08-图8 How Long Does It Take (25分)

题目地址

https://pta.patest.cn/pta/test/16/exam/4/question/674

5-12 How Long Does It Take   (25分)

Given the relations of all the activities of a project, you are supposed to find the earliest completion time of the project.

Input Specification:

Each input file contains one test case. Each case starts with a line containing two positive integers NN (le 100≤100), the number of activity check points (hence it is assumed that the check points are numbered from 0 to N-1N−1), and MM, the number of activities. Then MM lines follow, each gives the description of an activity. For the i-th activity, three non-negative numbers are given: S[i], E[i], and L[i], where S[i] is the index of the starting check point, E[i] of the ending check point, and L[i] the lasting time of the activity. The numbers in a line are separated by a space.

Output Specification:

For each test case, if the scheduling is possible, print in a line its earliest completion time; or simply output "Impossible".

Sample Input 1:

9 12
0 1 6
0 2 4
0 3 5
1 4 1
2 4 1
3 5 2
5 4 0
4 6 9
4 7 7
5 7 4
6 8 2
7 8 4

Sample Output 1:

18

Sample Input 2:

4 5
0 1 1
0 2 2
2 1 3
1 3 4
3 2 5

Sample Output 2:

Impossible

/*
评测结果
时间	结果	得分	题目	编译器	用时（ms）	内存（MB）	用户
2017-07-05 14:42	答案正确	25	5-12	gcc	18	1	
测试点结果
测试点	结果	得分/满分	用时（ms）	内存（MB）
测试点1	答案正确	15/15	2	1
测试点2	答案正确	2/2	2	1
测试点3	答案正确	4/4	2	1
测试点4	答案正确	2/2	2	1
测试点5	答案正确	2/2	18	1

简单的拓扑排序，注意处理多起点和多终点的问题
*/
#include<stdio.h>
#define MAXN 100
#define TRUE 1
#define FALSE 0
#define ERROR -1
struct checkpoint{
	int minStratTime;
	int inEdgeCount;
	int finished;
}gCheckpointTable[MAXN];

int gMatrix[MAXN][MAXN];

void InitCheckpointTable()
{
	int i;
	for(i=0;i<MAXN;i++)
	{
		gCheckpointTable[i].minStratTime=0;
		gCheckpointTable[i].inEdgeCount=0;
		gCheckpointTable[i].finished=0;
	}
}
void InitMatrix(N)
{
	int i,j;
	for(i=0;i<N;i++)
		for(j=0;j<N;j++)
			gMatrix[i][j]=ERROR;
}

int FindPoint(N) //返回一个入度为0并且没被访问过的点
{
	int i;
	int p=ERROR;
	for(i=0;i<N;i++)
	{
		if(gCheckpointTable[i].inEdgeCount==FALSE && gCheckpointTable[i].finished==FALSE)
			p=i;
	}
	return p;
}

void Calc(int N)
{
	int i,maxtime,p;
	while( (p=FindPoint(N)) != ERROR ) //如果还能返回入度为0的点 先叫它v1
	{
		gCheckpointTable[p].finished=TRUE; //先设为处理完成
		for(i=0;i<N;i++)//遍历该点所有发出去的边，找一个连接到的点v2
		{
			if(gMatrix[p][i]==ERROR)
				continue;
   			//如果被指向的结点v2，最小完成时间小于此节点v1的最小完成时间 加上  v1到v2的耗时，那么更新v2的最小时间
			if(gCheckpointTable[i].minStratTime<gCheckpointTable[p].minStratTime+gMatrix[p][i])
				gCheckpointTable[i].minStratTime = gCheckpointTable[p].minStratTime + gMatrix[p][i];
			gCheckpointTable[i].inEdgeCount--; //将v2的入度减一
		}
	}
	maxtime=0;
	for(i=0;i<N;i++) //考虑到有多终点问题，算完后把所有节点全扫一遍，挑一个结束时间最大的打出来
	{
		if(gCheckpointTable[i].finished==FALSE)
		{
			printf("Impossible");
			return;
		}
		if(gCheckpointTable[i].minStratTime>maxtime)
			maxtime=gCheckpointTable[i].minStratTime;
	}
	printf("%d",maxtime);
}

int main()
{
	int i;
	int N,M;
	int v1,v2,lastingTime;
	scanf("%d %d",&N,&M);
	InitMatrix(N);
	InitCheckpointTable();
	for(i=0;i<M;i++)
	{
		scanf("%d %d %d",&v1,&v2,&lastingTime);
		gMatrix[v1][v2]=lastingTime;
		gCheckpointTable[v2].inEdgeCount++;
	}
	Calc(N);
}