1001 难度 1 输入问题，还不能解决如何使得input的参数个数随意改变，只能先用vector动态存储，算法很简单
#include <cstdio>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <vector>
using namespace std;
#define MAX 10
int zero_count(int m, int n, int a[][MAX], int b[][MAX], int valid_column[])
{
int count = 0;
bool valid_row_flag = true;
for (int i = 0; i < m; i++)
{
int j;
for (j = 0; j < n; j++)
if (a[i][j] + b[i][j] != 0)
{
valid_column[j] = 0; //this is a invalid row
valid_row_flag = false;
}
if (valid_row_flag)
count++;
}
for (int j = 0; j < n; j++)
if (valid_column[j] != 0)
count++;
return count;
}
int main()
{
int a[MAX][MAX];
int b[MAX][MAX];
int valid_column[MAX];
int m, n;
vector<int> parameters;
int num;
while (scanf("%d", &num) != EOF)
{
parameters.push_back(num);
}
while (parameters[0] != 0)
{
m = parameters[0];
n = parameters[1];
memset(valid_column, 1, MAX * sizeof (int));
for (int i = 0; i < m; i++)
{
for (int j = 0; j < n; j++)
scanf("%d", &a[i][j]);
}
for (int i = 0; i < m; i++)
{
for (int j = 0; j < n; j++)
scanf("%d", &b[i][j]);
}
printf("%d", zero_count(m, n, a, b, valid_column));
printf("\n");
}
}
1005 结构体 用rear很巧妙
#include <iostream>
#include <algorithm>
using namespace std;
/*定义学生类，class和struct其实一个样，可以直接变为结构体*/
class Student
{
public:
int id;
int ge;
int gi;
int choice[5];
};
/*定义学校类*/
class School
{
public:
int quota;
Student std[100];
int front,rear;
};
/*学生成绩排序函数*/
bool cmp(Student a,Student b)
{
if(a.ge+a.gi!=b.ge+b.gi )
return a.ge+a.gi>b.ge+b.gi;
else
return a.ge>b.ge;
}
/*学生id排序函数*/
bool cmpa(Student a,Student b)
{
return a.id<b.id;
}
int main()
{
int n,m,k;
int i,j;
while(cin>>n>>m>>k)
{
Student std[100];
School sch[100];
/*学校信息录入*/
for(i=0;i!=m;i++)
{
cin>>sch[i].quota;
sch[i].front=0;
sch[i].rear=0;
}
/*学生信息录入*/
for(i=0;i!=n;i++)
{
cin>>std[i].ge>>std[i].gi;
for(j=0;j!=k;j++)
cin>>std[i].choice[j];
std[i].id=i;
}
/*学生成绩排序*/
sort(std,std+n,cmp);
/*按学生成绩从高到低分配学校*/
for(i=0;i!=n;i++)
{
/*按志愿先后选择学校*/
for(j=0;j!=k;j++)
{
/*学校还有名额时，直接入该学校的队列*/
if(sch[std[i].choice[j]].quota>0)
{
sch[std[i].choice[j]].std[sch[std[i].choice[j]].rear++]=std[i];
sch[std[i].choice[j]].quota--;
break;
}
/*学校没有名额时，成绩是否与最后一名录取的学生相同，同则进*/
else if(std[i].ge==sch[std[i].choice[j]].std[sch[std[i].choice[j]].rear-1].ge
&&std[i].gi==sch[std[i].choice[j]].std[sch[std[i].choice[j]].rear-1].gi)
{
sch[std[i].choice[j]].std[sch[std[i].choice[j]].rear++]=std[i];
sch[std[i].choice[j]].quota--;
break;
}
}
}
/*输出每个学校录取者的id*/
for(i=0;i!=m;i++)
{
if(sch[i].front==sch[i].rear)
cout<<endl;
else
{
/*学生id排序*/
sort(sch[i].std,sch[i].std+sch[i].rear,cmpa);
cout<<sch[i].std[sch[i].front++].id;
while(sch[i].front!=sch[i].rear)
cout<<' '<<sch[i].std[sch[i].front++].id;
cout<<endl;
}
}
}
return 0;
}1008 最短路径， dijkstra算法，难度1，但是要注意在每次使用指针时都要检查，尤其是在循环条件中，开始循环，和开始下一次循环都要注意，使用adj，用节点作为关键字索引
#include <iostream>
#define MAX_NODE 1000
#include <vector>
#include <fstream>
#define INF 100
using namespace std;
struct Edge
{
int fr_id, to_id;
int cost, dist;
int dist_source;
Edge *next;
Edge(int fr, int to, int co, int di, int di_so, Edge* n)
{
fr_id = fr;
to_id = to;
dist = di;
cost = co;
dist_source = di_so;
next = n;
}
};
struct Node
{
int id;
Edge *firstEdge;
};
int short_path(vector<Edge*> v, vector<int> &pv, int s, int d)
{
Edge *p = v[s]; //find the first edge of source
int min = p->dist;
int index = p->to_id;
while (p)
{
if (p->dist < min)
{
min = p->dist;
index = p->to_id;
}
p = p->next; //find the shortest path from source
}
int distance = min; //distance now
pv[index] = s;
//record index's parent
while (1)
{
if (index == d) //check if it is the destination
break;
p = v[index]; //find the shortest pathe from index
if (p == 0)
break;
min = p->dist_source;
int min_index = p->to_id;
while (p)
{
//find the less dist to source
if (p->dist + distance < p->dist_source) //update each node's distance to the source, for we don't know which is the end
{
p->dist_source = p->dist + distance;
if (p->dist_source < min || (p->dist_source == min && p->cost < v[min_index]->cost))
{
min = p->dist_source;
min_index = p->to_id;
}
}
p = p->next;
}
pv[min_index] = index;
index = min_index;
}
return v[pv[d]]->dist_source;
}
int main()
{
int nodes, edges;
if(cin >> nodes >> edges)
{
//if (nodes == 0 && edges == 0)
//break;
vector<Edge*> a(nodes);
vector<int> parent(nodes);
int from, to, dist, cost;
while (edges--)
{
cin >> from >> to >> dist >> cost;
if (a[from] == 0)
{
Edge *edge = new Edge(from, to, cost, dist, INF, NULL);
a[from] = edge;
}
else
{
Edge *edge = new Edge(from, to, cost, dist, INF, a[from]);
a[from] = edge;
}
}
int sour, dest;
cin >> sour >> dest;
cout << "shortest path is " << short_path(a, parent, sour, dest) << endl;
int pp = parent[dest];
cout << dest;
while (pp != sour)
{
cout << "<-" << pp;
pp = parent[pp];
}
cout << "<-" << sour << endl;
}
}