poj 3159 Candies(差分制約)
15223 ワード
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Candies
Time Limit: 1500MS
Memory Limit: 131072K
Description
During the kindergarten days, flymouse was the monitor of his class. Occasionally the head-teacher brought the kids of flymouse’s class a large bag of candies and had flymouse distribute them. All the kids loved candies very much and often compared the numbers of candies they got with others. A kid A could had the idea that though it might be the case that another kid B was better than him in some aspect and therefore had a reason for deserving more candies than he did, he should never get a certain number of candies fewer than B did no matter how many candies he actually got, otherwise he would feel dissatisfied and go to the head-teacher to complain about flymouse’s biased distribution.
snoopy shared class with flymouse at that time. flymouse always compared the number of his candies with that of snoopy’s. He wanted to make the difference between the numbers as large as possible while keeping every kid satisfied. Now he had just got another bag of candies from the head-teacher, what was the largest difference he could make out of it?
Input
The input contains a single test cases. The test cases starts with a line with two integers N and M not exceeding 30 000 and 150 000 respectively. N is the number of kids in the class and the kids were numbered 1 through N. snoopy and flymouse were always numbered 1 and N. Then follow M lines each holding three integers A, B and c in order, meaning that kid A believed that kid B should never get over c candies more than he did.
Output
Output one line with only the largest difference desired. The difference is guaranteed to be finite.
Sample Input
Sample Output
Hint
32-bit signed integer type is capable of doing all arithmetic.
タイトル:
nの個人番号は1からn、mの要求があり、各要求はui、vi、wiである.代表番号uiの人が分けたキャンディは、viの人よりwi個少ないことが多い.nの番号を求める人は、1の番号の人よりも多くのキャンディを求めることができます.
分析:
x[i]代表番号iの人が分けたキャンディの数を設定します.
次式x[vi]−x[ui]<=wiが得られる.
この式によれば、建図はuiからviへの重み値がwiの有向辺である.
その後,最短ルートを用いて解く.
注意:この問題spfaではキューでTLEができる場合、スタックはACができます.もちろん,この問題の重み値はすべて正であるためdijkstraを用いることができる.
コード君
Candies
Time Limit: 1500MS
Memory Limit: 131072K
Description
During the kindergarten days, flymouse was the monitor of his class. Occasionally the head-teacher brought the kids of flymouse’s class a large bag of candies and had flymouse distribute them. All the kids loved candies very much and often compared the numbers of candies they got with others. A kid A could had the idea that though it might be the case that another kid B was better than him in some aspect and therefore had a reason for deserving more candies than he did, he should never get a certain number of candies fewer than B did no matter how many candies he actually got, otherwise he would feel dissatisfied and go to the head-teacher to complain about flymouse’s biased distribution.
snoopy shared class with flymouse at that time. flymouse always compared the number of his candies with that of snoopy’s. He wanted to make the difference between the numbers as large as possible while keeping every kid satisfied. Now he had just got another bag of candies from the head-teacher, what was the largest difference he could make out of it?
Input
The input contains a single test cases. The test cases starts with a line with two integers N and M not exceeding 30 000 and 150 000 respectively. N is the number of kids in the class and the kids were numbered 1 through N. snoopy and flymouse were always numbered 1 and N. Then follow M lines each holding three integers A, B and c in order, meaning that kid A believed that kid B should never get over c candies more than he did.
Output
Output one line with only the largest difference desired. The difference is guaranteed to be finite.
Sample Input
2 2
1 2 5
2 1 4Sample Output
5Hint
32-bit signed integer type is capable of doing all arithmetic.
タイトル:
nの個人番号は1からn、mの要求があり、各要求はui、vi、wiである.代表番号uiの人が分けたキャンディは、viの人よりwi個少ないことが多い.nの番号を求める人は、1の番号の人よりも多くのキャンディを求めることができます.
分析:
x[i]代表番号iの人が分けたキャンディの数を設定します.
次式x[vi]−x[ui]<=wiが得られる.
この式によれば、建図はuiからviへの重み値がwiの有向辺である.
その後,最短ルートを用いて解く.
注意:この問題spfaではキューでTLEができる場合、スタックはACができます.もちろん,この問題の重み値はすべて正であるためdijkstraを用いることができる.
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <queue>
5 #include <stack>
6 #include <cstdlib>
7 using namespace std;
8 #define REP(A,X) for(int A=0;A<X;A++)
9 #define MAXE 200010
10 #define MAXP 30010
11 #define INF 0x7fffffff
12 #define MP(A,B) make_pair(A,B)
13 typedef pair<int,int> PII ;
14 struct node{
15 int v,d,next;
16 }edge[MAXE];
17 int e=0;
18 int head[MAXP];
19 void init(){
20 e=0;
21 REP(i,MAXP)head[i]=-1;
22 }
23 void add_edge(int u,int v,int d)
24 {
25 edge[e].v=v;
26 edge[e].d=d;
27 edge[e].next=head[u];
28 head[u]=e;
29 e++;
30 }
31 int vis[MAXP],dis[MAXP];
32 void spfa()
33 {
34 REP(i,MAXP)vis[i]=0;
35 REP(i,MAXP)dis[i]=i==1?0:INF;
36 //queue<int>q;
37 stack<int>q;
38 q.push(1);
39 vis[1]=1;
40 while(!q.empty()){
41 //int x=q.front();
42 int x=q.top();
43 q.pop();
44 for(int t=head[x];t!=-1;t=edge[t].next)
45 {
46 int y=edge[t].v;
47 int d=edge[t].d;
48 if(dis[y]>dis[x]+d){
49 dis[y]=dis[x]+d;
50 if(!vis[y]){
51 q.push(y);
52 vis[y]=1;
53 }
54 }
55 }
56 vis[x]=0;
57 }
58 }
59 void dijkstra(int s)
60 {
61 REP(i,MAXP)vis[i]=0;
62 REP(i,MAXP)dis[i]=i==s?0:INF;
63 priority_queue<PII,vector<PII>,greater<PII> > q;
64 q.push(MP(dis[s],s));
65 while(!q.empty())
66 {
67 PII p=q.top();
68 q.pop();
69 int x=p.second;
70 if(vis[x])continue;
71 vis[x]=1;
72 for(int t=head[x];t!=-1;t=edge[t].next)
73 {
74 int y=edge[t].v;
75 int d=edge[t].d;
76 if(!vis[y]&&dis[y]>dis[x]+d)
77 {
78 dis[y]=dis[x]+d;
79 q.push(MP(dis[y],y));
80 }
81 }
82 }
83
84 }
85
86 int main()
87 {
88 int m,n;
89 while(scanf("%d%d",&n,&m)!=EOF){
90 int u,v,w;
91 init();
92 REP(i,m){
93 scanf("%d%d%d",&u,&v,&w);
94 add_edge(u,v,w);
95 }
96 //dijkstra(1);
97 spfa();
98 printf("%d
",dis[n]);
99 }
100 return 0;
101 }コード君