A：
题干：
Input
The first line contains integer n (1 ≤ n ≤ 100) — the number of cars.
Each of the next n lines contains n space-separated integers that determine matrix A.
It is guaranteed that on the main diagonal there are - 1, and - 1 doesn't appear anywhere else in the matrix.
It is guaranteed that the input is correct, that is, if Aij = 1, then Aji = 2, if Aij = 3, then Aji = 3, and if Aij = 0, then Aji = 0.
Output
Print the number of good cars and in the next line print their space-separated indices in the increasing order.
Examples
Input
3 -1 0 0 0 -1 1 0 2 -1
Output
2 1 3
Input
4 -1 3 3 3 3 -1 3 3 3 3 -1 3 3 3 3 -1
Output
0
题目大意：
给定一个N*N的矩阵a来描述N辆车的关系，如果a(i,j)=0代表这两辆车都没翻车，=1代表i车翻车了，=2代表j车翻车了，=3代表都翻车了。问你N辆车中有多少车是完好无损的？
解题报告：
模拟、、
AC代码：
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
const int MAX = 2e5 + 5;
int n;
int maze[555][555],qq[MAX];
int main()
{
cin>>n;
for(int i = 1; i<=n; i++) {
for(int j = 1; j<=n; j++) {
scanf("%d",&maze[i][j]);
}
}
int ans = 0;
for(int i = 1;i<=n; i++) {
bool flag = 1;
for(int j = 1; j<=n; j++) {
if(i == j) continue;
if(maze[i][j] == 1 || maze[i][j] == 3) flag = 0;
}
if(flag) ++ans,qq[ans] = i;
}
printf("%d\n",ans);
for(int i = 1; i<=ans; i++) {
printf("%d%c",qq[i],i == ans ? '\n' : ' ');
}
return 0 ;
}B：
题干：
Little Susie loves strings. Today she calculates distances between them. As Susie is a small girl after all, her strings contain only digits zero and one. She uses the definition of Hamming distance:
We will define the distance between two strings s and t of the same length consisting of digits zero and one as the number of positions i, such that si isn't equal to ti.
As besides everything else Susie loves symmetry, she wants to find for two strings sand t of length n such string p of length n, that the distance from p to s was equal to the distance from p to t.
It's time for Susie to go to bed, help her find such string p or state that it is impossible.
Input
The first line contains string s of length n.
The second line contains string t of length n.
The length of string n is within range from 1 to 105. It is guaranteed that both strings contain only digits zero and one.
Output
Print a string of length n, consisting of digits zero and one, that meets the problem statement. If no such string exist, print on a single line "impossible" (without the quotes).
If there are multiple possible answers, print any of them.
Examples
Input
0001 1011
Output
0011
Input
000 111
Output
impossible
Note
In the first sample different answers are possible, namely — 0010, 0011, 0110, 0111, 1000, 1001, 1100, 1101.
题目大意：
先给出了汉明距离的定义，然后输入s，t两个字符串。让你构造一个字符串p，st. Hamming(s,p) == Hamming(p.t)。
解题报告：
贪心就行了、
AC代码：
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
const int MAX = 2e5 + 5;
char s[MAX],t[MAX];
int main()
{
cin>>(s+1);
cin>>(t+1);
int qq = 0;
int len = strlen(s+1);
for(int i = 1; i<=len; i++) {
if(s[i] != t[i]) qq++;
}
if(qq%2 == 1) puts("impossible");
else {
int cur = 0;
for(int i = 1; i<=len; i++) {
if(s[i] != t[i] && cur*2 < qq) {
cur++;printf("%c",t[i]);
}
else printf("%c",s[i]);
}
}
return 0 ;
}C：
题干：
Little Susie listens to fairy tales before bed every day. Today's fairy tale was about wood cutters and the little girl immediately started imagining the choppers cutting wood. She imagined the situation that is described below.
There are n trees located along the road at points with coordinates x1, x2, ..., xn. Each tree has its height hi. Woodcutters can cut down a tree and fell it to the left or to the right. After that it occupies one of the segments [xi - hi, xi] or [xi;xi + hi]. The tree that is not cut down occupies a single point with coordinate xi. Woodcutters can fell a tree if the segment to be occupied by the fallen tree doesn't contain any occupied point. The woodcutters want to process as many trees as possible, so Susie wonders, what is the maximum number of trees to fell.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of trees.
Next n lines contain pairs of integers xi, hi (1 ≤ xi, hi ≤ 109) — the coordinate and the height of the і-th tree.
The pairs are given in the order of ascending xi. No two trees are located at the point with the same coordinate.
Output
Print a single number — the maximum number of trees that you can cut down by the given rules.
Examples
Input
5 1 2 2 1 5 10 10 9 19 1
Output
3
Input
5 1 2 2 1 5 10 10 9 20 1
Output
4
Note
In the first sample you can fell the trees like that:
- fell the 1-st tree to the left — now it occupies segment [ - 1;1]
- fell the 2-nd tree to the right — now it occupies segment [2;3]
- leave the 3-rd tree — it occupies point 5
- leave the 4-th tree — it occupies point 10
- fell the 5-th tree to the right — now it occupies segment [19;20]
In the second sample you can also fell 4-th tree to the right, after that it will occupy segment [10;19].
题目大意：
无限长的数轴上有n个点，每个点的坐标为x[i]，种有高度为h[i]的树，现在要把一些树砍到，被砍倒的树要么倒向左边，要么倒向右边，会分别把[xi - hi, xi] 和 [xi,xi + hi]占用，如果某棵树不被砍倒，那么它就只占用x[i]这一个点的位置，现在给定你n个点的x[i],h[i]，问最多能砍倒几棵树？（n<=1e5, x[i],h[i]<=1e9）
解题报告：
多阶段决策问题，显然dp。（不过这题好像贪心也能解决？因为关于第i棵树的摆放，看他能不能往左倒，能倒就倒，否则看他如果往右倒那后面那颗能不能往左倒，，如果同时可以那就可以往右倒，如果左边这个不可以倒，那就看右边，如果左边这个可以倒，那就一定倒。（为什么呢？因为你可以倒，那么就算你不倒，右边那个如果往左倒，答案相同且对后面没有影响，如果直立，答案还不如这种选择，如果往右倒，那更不如这种选择（因为这段空间就空着了啊）），，这样贪心就行了，，但是显然没有dp简单。）
AC代码：
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
const int MAX = 2e5 + 5;
int n;
ll x[MAX],h[MAX];
ll dp[MAX][3];//状态表示：0不倒 1左边 2右边
ll Max(ll a,ll b,ll c) {
return max(a,max(b,c));
}
int main()
{
cin>>n;
for(int i = 1; i<=n; i++) scanf("%lld %lld",x+i,h+i);
x[0] = -0x3f3f3f3f;
x[n+1] = 0x3f3f3f3f3f3f;
ll ans = 0;
for(int i = 1; i<=n; i++) {
dp[i][0] = Max(dp[i-1][0],dp[i-1][1],dp[i-1][2]);
if(x[i+1] - x[i] > h[i]) dp[i][2] = dp[i][0] + 1;
else dp[i][2] = 0;
dp[i][1] = max(dp[i-1][0],dp[i-1][2]);
if(x[i] - x[i-1] > h[i]) dp[i][1] = max(dp[i][1],max(dp[i-1][1],dp[i-1][0])+1);
if(x[i] - x[i-1] > h[i] + h[i-1]) dp[i][1] = max(dp[i][1],dp[i-1][2]+1);
}
printf("%lld\n",Max(dp[n][0],dp[n][1],dp[n][2]));
return 0 ;
}D：
题干：
Little girl Susie went shopping with her mom and she wondered how to improve service quality.
There are n people in the queue. For each person we know time ti needed to serve him. A person will be disappointed if the time he waits is more than the time needed to serve him. The time a person waits is the total time when all the people who stand in the queue in front of him are served. Susie thought that if we swap some people in the queue, then we can decrease the number of people who are disappointed.
Help Susie find out what is the maximum number of not disappointed people can be achieved by swapping people in the queue.
Input
The first line contains integer n (1 ≤ n ≤ 105).
The next line contains n integers ti (1 ≤ ti ≤ 109), separated by spaces.
Output
Print a single number — the maximum number of not disappointed people in the queue.
Examples
Input
5 15 2 1 5 3
Output
4
Note
Value 4 is achieved at such an arrangement, for example: 1, 2, 3, 5, 15. Thus, you can make everything feel not disappointed except for the person with time 5.
题目大意：
有n个人，每个人都有一个等待时间，如果对于当前的人来说总等待时间超过自己的等待时间，那么这个人就会失望，问换一下顺序，使失望的人最少，问最多有多少个人不失望。
解题报告：
贪心，首先根据直觉排序，，然后如果他不失望那更好，如果他失望了，那么反正他也得失望，我们还不如把他换到最后去让他失望的更多一些，所以扫一遍就行了。
AC代码：
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
const int MAX = 2e5 + 5;
int n;
ll a[MAX];
int main()
{
cin>>n;
for(int i = 1; i<=n; i++) scanf("%lld",a+i);
sort(a+1,a+n+1);
ll cur = 0,ans = 0;
for(int i = 1; i<=n; i++) {
if(cur <= a[i]) {
ans++;cur += a[i];
}
}
printf("%lld\n",ans);
return 0 ;
}E：
题干：
Little girl Susie accidentally found her elder brother's notebook. She has many things to do, more important than solving problems, but she found this problem too interesting, so she wanted to know its solution and decided to ask you about it. So, the problem statement is as follows.
Let's assume that we are given a connected weighted undirected graph G = (V, E) (here V is the set of vertices, E is the set of edges). The shortest-path tree from vertex u is such graph G1 = (V, E1) that is a tree with the set of edges E1 that is the subset of the set of edges of the initial graph E, and the lengths of the shortest paths from u to any vertex to G and to G1 are the same.
You are given a connected weighted undirected graph G and vertex u. Your task is to find the shortest-path tree of the given graph from vertex u, the total weight of whose edges is minimum possible.
Input
The first line contains two numbers, n and m (1 ≤ n ≤ 3·105, 0 ≤ m ≤ 3·105) — the number of vertices and edges of the graph, respectively.
Next m lines contain three integers each, representing an edge — ui, vi, wi — the numbers of vertices connected by an edge and the weight of the edge (ui ≠ vi, 1 ≤ wi ≤ 109). It is guaranteed that graph is connected and that there is no more than one edge between any pair of vertices.
The last line of the input contains integer u (1 ≤ u ≤ n) — the number of the start vertex.
Output
In the first line print the minimum total weight of the edges of the tree.
In the next line print the indices of the edges that are included in the tree, separated by spaces. The edges are numbered starting from 1 in the order they follow in the input. You may print the numbers of the edges in any order.
If there are multiple answers, print any of them.
Examples
Input
3 3 1 2 1 2 3 1 1 3 2 3
Output
2 1 2
Input
4 4 1 2 1 2 3 1 3 4 1 4 1 2 4
Output
4 2 3 4
Note
In the first sample there are two possible shortest path trees:
- with edges 1 – 3 and 2 – 3 (the total weight is 3);
- with edges 1 – 2 and 2 – 3 (the total weight is 2);
And, for example, a tree with edges 1 – 2 and 1 – 3 won't be a shortest path tree for vertex 3, because the distance from vertex 3 to vertex 2 in this tree equals 3, and in the original graph it is 1.
题目大意：
给定一个n个结点m条边的无向图，并给出源点s，让你找出图中权值最小的最短路树，并输出这个权值，和选择的边的编号。
定义最短路树：设最短路的图为图G，最短路树的图为图Q，则源点s到G中任意一个顶点的距离，应该等于s到Q中任意一个顶点的距离，且Q为一棵树。
解题报告：
首先不难证明，这个其实就是在所有最短路的可能的边中去选择其中的一些，其实可以看成是松弛（将一个点松弛成两个点，也就是将边数变得越多越好，当然，这是在保证最短路的基础之上的）。所以我们可以在Dijkstra的时候同时维护边权的最小值（也就是如果dis[] > dis[] + e[i].w 更新，else if dis[] == dis[] + e[i].w，则取权值最小，并且将该边的编号记录到对应的e[i].to对应的数组上，Dijkstra完了之后就结束了，最后按要求输出就行了）。
这题我选择了另外一种做法，先跑一边裸的Dijkstra，然后我们得到了最短路数组dis，然后枚举除了起点之外的每一个点，然后枚举他的每一条边，并记录它到起点的那条边的边权最小值，这条边一定选在这棵树内，最后按要求输出。这样写代码更好实现一些。其实两个方法是等价的。
AC代码：
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#include<ctime>
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
const int MAX = 6e5 + 5;
struct Edge {
int to,ne,ii;
ll w;
} e[MAX];
int head[MAX],tot,n,m;
bool vis[MAX];
int ans[MAX],cnt;
ll dis[MAX];
void add(int u,int v,ll w,int i) {
e[++tot].to = v;
e[tot].ne = head[u];
e[tot].w = w;
e[tot].ii = i;
head[u] = tot;
}
struct Point {
int id;
ll c;
Point(){}
Point(int id,ll c):id(id),c(c){}
bool operator<(const Point & b) const{
return c > b.c;
}
};
void Dijkstra(int st) {
for(int i = 1; i<=n; i++) dis[i] = LLONG_MAX;
dis[st] = 0;
priority_queue<Point> pq;
pq.push(Point(st,0));
while(!pq.empty()) {
Point cur = pq.top();pq.pop();
if(vis[cur.id]) continue;
vis[cur.id] = 1;
for(int i = head[cur.id]; i!=-1; i = e[i].ne) {
int v = e[i].to;
if(dis[v] > dis[cur.id] + e[i].w) {
dis[v] = dis[cur.id] + e[i].w;
pq.push(Point(v,dis[v]));
}
}
}
}
int main()
{
int u,v;
memset(head,-1,sizeof head);
ll w;
cin>>n>>m;
for(int i = 1; i<=m; i++) {
scanf("%d%d%lld",&u,&v,&w);
add(u,v,w,i);
add(v,u,w,i);
}
cin>>u;
Dijkstra(u);
int minid;
ll ANS = 0;
for(int i = 1; i<=n; i++) {
if(i == u) continue;
ll minn = LLONG_MAX;
for(int j = head[i]; j != -1; j = e[j].ne) {
if(dis[i] == dis[e[j].to] + e[j].w && minn > e[j].w) {
minn = e[j].w;
minid = e[j].ii;
}
}
ANS += minn;
ans[++cnt] = minid;
}
printf("%lld\n",ANS);
for(int i = 1; i<=cnt; i++) {
printf("%d ",ans[i]);
}
return 0 ;
}