題目鏈接：https://vjudge.net/problem/POJ-3258

題目大意

　　給定數(shù)軸上一個(gè)起點(diǎn) 0，終點(diǎn) L，以及中間 N 個(gè)不同的點(diǎn)，現(xiàn)準(zhǔn)備刪除中間 N 個(gè)點(diǎn)中的 M 個(gè)，使得剩下來(lái)的點(diǎn)（包括起點(diǎn)和終點(diǎn)），相鄰兩點(diǎn)間距的最小值最大，求這個(gè)值。

分析

二分最短距離即可。

代碼如下

1 #include <cmath> 2 #include <ctime> 3 #include <iostream> 4 #include <string> 5 #include <vector> 6 #include <cstdio> 7 #include <cstdlib> 8 #include <cstring> 9 #include <queue> 10 #include <map> 11 #include <set> 12 #include <algorithm> 13 #include <cctype> 14 #include <stack> 15 #include <deque> 16 #include <list> 17 #include <sstream> 18 #include <cassert> 19 using namespace std; 20 21 #define INIT() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0); 22 #define Rep(i,n) for (int i = 0; i < (n); ++i) 23 #define For(i,s,t) for (int i = (s); i <= (t); ++i) 24 #define rFor(i,t,s) for (int i = (t); i >= (s); --i) 25 #define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i) 26 #define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i) 27 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i) 28 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i) 29 30 #define pr(x) cout << #x << " = " << x << " " 31 #define prln(x) cout << #x << " = " << x << endl 32 33 #define LOWBIT(x) ((x)&(-x)) 34 35 #define ALL(x) x.begin(),x.end() 36 #define INS(x) inserter(x,x.begin()) 37 #define UNIQUE(x) x.erase(unique(x.begin(), x.end()), x.end()) 38 #define REMOVE(x, c) x.erase(remove(x.begin(), x.end(), c), x.end()); // 刪去 x 中所有 c 39 #define TOLOWER(x) transform(x.begin(), x.end(), x.begin(),::tolower); 40 #define TOUPPER(x) transform(x.begin(), x.end(), x.begin(),::toupper); 41 42 #define ms0(a) memset(a,0,sizeof(a)) 43 #define msI(a) memset(a,inf,sizeof(a)) 44 #define msM(a) memset(a,-1,sizeof(a)) 45 46 #define MP make_pair 47 #define PB push_back 48 #define ft first 49 #define sd second 50 51 template<typename T1, typename T2> 52 istream &operator>>(istream &in, pair<T1, T2> &p) { 53 in >> p.first >> p.second; 54 return in; 55 } 56 57 template<typename T> 58 istream &operator>>(istream &in, vector<T> &v) { 59 for (auto &x: v) 60 in >> x; 61 return in; 62 } 63 64 template<typename T1, typename T2> 65 ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) { 66 out << "[" << p.first << ", " << p.second << "]" << "\n"; 67 return out; 68 } 69 70 inline int gc(){ 71 static const int BUF = 1e7; 72 static char buf[BUF], *bg = buf + BUF, *ed = bg; 73 74 if(bg == ed) fread(bg = buf, 1, BUF, stdin); 75 return *bg++; 76 } 77 78 inline int ri(){ 79 int x = 0, f = 1, c = gc(); 80 for(; c<48||c>57; f = c=='-'?-1:f, c=gc()); 81 for(; c>47&&c<58; x = x*10 + c - 48, c=gc()); 82 return x*f; 83 } 84 85 template<class T> 86 inline string toString(T x) { 87 ostringstream sout; 88 sout << x; 89 return sout.str(); 90 } 91 92 inline int toInt(string s) { 93 int v; 94 istringstream sin(s); 95 sin >> v; 96 return v; 97 } 98 99 //min <= aim <= max 100 template<typename T> 101 inline bool BETWEEN(const T aim, const T min, const T max) { 102 return min <= aim && aim <= max; 103 } 104 105 typedef long long LL; 106 typedef unsigned long long uLL; 107 typedef pair< double, double > PDD; 108 typedef pair< int, int > PII; 109 typedef pair< int, PII > PIPII; 110 typedef pair< string, int > PSI; 111 typedef pair< int, PSI > PIPSI; 112 typedef set< int > SI; 113 typedef set< PII > SPII; 114 typedef vector< int > VI; 115 typedef vector< double > VD; 116 typedef vector< VI > VVI; 117 typedef vector< SI > VSI; 118 typedef vector< PII > VPII; 119 typedef map< int, int > MII; 120 typedef map< LL, int > MLLI; 121 typedef map< int, string > MIS; 122 typedef map< int, PII > MIPII; 123 typedef map< PII, int > MPIII; 124 typedef map< string, int > MSI; 125 typedef map< string, string > MSS; 126 typedef map< PII, string > MPIIS; 127 typedef map< PII, PII > MPIIPII; 128 typedef multimap< int, int > MMII; 129 typedef multimap< string, int > MMSI; 130 //typedef unordered_map< int, int > uMII; 131 typedef pair< LL, LL > PLL; 132 typedef vector< LL > VL; 133 typedef vector< VL > VVL; 134 typedef priority_queue< int > PQIMax; 135 typedef priority_queue< int, VI, greater< int > > PQIMin; 136 const double EPS = 1e-8; 137 const LL inf = 0x3fffffff; 138 const LL infLL = 0x3fffffffffffffffLL; 139 const LL mod = 1e9 + 7; 140 const int maxN = 1e5 + 7; 141 const LL ONE = 1; 142 const LL evenBits = 0xaaaaaaaaaaaaaaaa; 143 const LL oddBits = 0x5555555555555555; 144 145 int L, N, M; 146 int dis[maxN >> 1], l, r; 147 148 int main(){ 149 //freopen("MyOutput.txt","w",stdout); 150 //freopen("input.txt","r",stdin); 151 //INIT(); 152 while(~scanf("%d %d %d", &L, &N, &M)) { 153 dis[0] = 0; 154 For(i, 1, N) scanf("%d", &dis[i]); 155 dis[++N] = L; 156 sort(dis, dis + N + 1); 157 l = dis[1]; 158 r = L; 159 Rep(i, N) { 160 dis[i] = dis[i + 1] - dis[i]; 161 l = min(l, dis[i]); 162 } 163 164 while(l < r) { 165 int mid = (l + r) >> 1, cnt = 0, sum = 0; 166 167 Rep(i, N) { 168 sum += dis[i]; 169 if(sum > mid) sum = 0; // 這里看的是 mid + 1 170 else ++cnt; 171 } 172 173 if(cnt > M) r = mid; 174 else l = mid + 1; 175 } 176 177 printf("%d\n", l); 178 } 179 return 0; 180 } 181 182 /* 183 16 7 5 184 2 4 6 8 10 12 14 185 16 0 0 186 187 Ans: 188 4 189 16 190 */ View Code

?

轉(zhuǎn)載于:https://www.cnblogs.com/zaq19970105/p/11182735.html

創(chuàng)作挑戰(zhàn)賽新人創(chuàng)作獎(jiǎng)勵(lì)來(lái)咯，堅(jiān)持創(chuàng)作打卡瓜分現(xiàn)金大獎(jiǎng)

總結(jié)

以上是生活随笔為你收集整理的POJ 3258 River Hopscotch的全部?jī)?nèi)容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網(wǎng)站內(nèi)容還不錯(cuò)，歡迎將生活随笔推薦給好友。