2016 ACM-ICPC World Finals - Phuket
签到题，暴力
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int maxn = 1e5 + 10; 4 int rt[maxn], ls[maxn], rs[maxn], val[maxn], cnt; 5 6 void add(int x, int v) { 7 if(v < val[x]) { 8 if(ls[x]) add(ls[x], v); 9 else { 10 ls[x] = ++cnt; 11 val[cnt] = v; 12 return; 13 } 14 } 15 else { 16 if(rs[x]) add(rs[x], v); 17 else { 18 rs[x] = ++cnt; 19 val[cnt] = v; 20 return; 21 } 22 } 23 } 24 25 bool check(int x, int y) { 26 bool ret = true; 27 if(ls[x] && !ls[y] || !ls[x] && ls[y]) return false; 28 if(ls[x]) ret &= check(ls[x], ls[y]); 29 if(rs[x] && !rs[y] || !rs[x] && rs[y]) return false; 30 if(rs[x]) ret &= check(rs[x], rs[y]); 31 return ret; 32 } 33 34 int main() { 35 int n, k; 36 scanf("%d %d", &n, &k); 37 for(int i = 1; i <= n; ++i) { 38 int x; 39 scanf("%d", &x); 40 rt[i] = ++cnt, val[cnt] = x; 41 for(int j = 2; j <= k; ++j) { 42 scanf("%d", &x); 43 add(rt[i], x); 44 } 45 for(int j = 1; j < i; ++j) { 46 if(check(rt[i], rt[j])) { 47 i--; n--; break; 48 } 49 } 50 } 51 printf("%d\n", n); 52 return 0; 53 }
EOJ大鱼吃小鱼
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int maxn = 1e6 + 10; 4 typedef long long ll; 5 int a[maxn], b[maxn], id[maxn]; 6 7 bool cmp(int i, int j) { 8 if(a[i] <= b[i] && a[j] > b[j]) return true; 9 if(a[i] > b[i] && a[j] <= b[j]) return false; 10 if(a[i] <= b[i]) return a[i] < a[j]; 11 return b[i] > b[j]; 12 } 13 14 int main() { 15 int n; 16 scanf("%d", &n); 17 for(int i = 1; i <= n; ++i) { 18 scanf("%d %d", a + i, b + i); 19 id[i] = i; 20 } 21 sort(id + 1, id + 1 + n, cmp); 22 ll ex = 0, ans = 0; 23 for(int i = 1; i <= n; ++i) { 24 int o = id[i]; 25 if(a[o] > ex) ans += a[o] - ex, ex += a[o] - ex; 26 ex -= a[o] - b[o]; 27 } 28 printf("%lld\n", ans); 29 return 0; 30 }
枚举1e6内基数，否则不会超过四位可以解三次方程，系数都是正的单调可以二分
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 5 inline double dcal(int i, int j, int p, int q, ll x) { 6 double y = 0. + x; 7 return i * y * y * y + j * y * y + p * y + q; 8 } 9 10 inline ll cal(int i, int j, int p, int q, ll x) { 11 return i * x * x * x + j * x * x + p * x + q; 12 } 13 14 int main() { 15 ll y, l, ans = 10; 16 scanf("%lld %lld", &y, &l); 17 for(int i = 0; i <= 9; ++i) { 18 for(int j = 0; j <= 9; ++j) { 19 for(int p = 0; p <= 9; ++p) { 20 for(int q = 0; q <= 9; ++q) { 21 if(1000 * i + 100 * j + 10 * p + q < l) continue; 22 ll lb = 11, rb = 1e18; 23 while(lb < rb) { 24 ll mid = (lb + rb) / 2; 25 if(dcal(i, j, p, q, mid) > y + 1 || cal(i, j, p, q, mid) >= y) rb = mid; 26 else lb = mid + 1; 27 } 28 if(cal(i, j, p, q, rb) == y) ans = max(ans, rb); 29 } 30 } 31 } 32 } 33 for(ll b = 11; b <= 1000000; ++b) { 34 int ok = 1; 35 ll t = y, x = 1, s = 0; 36 while(t) { 37 ll r = t % b; 38 if(r >= 10) {ok = 0; break;} 39 t /= b; 40 s += x * r; 41 x *= 10; 42 } 43 if(ok && s >= l) ans = max(ans, b); 44 } 45 printf("%lld\n", ans); 46 return 0; 47 }
枚举端点极角排序扫一遍
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 const int maxn = 2222; 5 ll x[2][maxn], y[maxn]; 6 int vis[maxn]; 7 8 struct point { 9 ll x, y, id; 10 point(ll _x = 0, ll _y = 0, ll _id = 0) { 11 x = _x, y = _y, id = _id; 12 } 13 }; 14 15 bool cmp(point A, point B) { 16 return A.x * B.y > A.y * B.x; 17 } 18 19 int main() { 20 int n; 21 scanf("%d", &n); 22 for(int i = 1; i <= n; ++i) { 23 scanf("%lld %lld %lld", x[0] + i, x[1] + i, y + i); 24 if(x[0][i] > x[1][i]) swap(x[0][i], x[1][i]); 25 } 26 ll ans = 0; 27 for(int i = 1; i <= n; ++i) { 28 for(int oi = 0; oi < 2; ++oi) { 29 ll xo = x[oi][i], yo = y[i]; 30 vector<point> p; 31 for(int j = 1; j <= n; ++j) { 32 if(j == i) continue; 33 for(int oj = 0; oj < 2; ++oj) { 34 ll xj = x[oj][j] - xo, yj = y[j] - yo; 35 if(yj == 0) continue; 36 if(yj < 0) xj = -xj, yj = -yj; 37 p.emplace_back(point(xj, yj, j)); 38 } 39 } 40 sort(p.begin(), p.end(), cmp); 41 memset(vis, 0, sizeof(vis)); 42 int sz = int(p.size()); 43 ll tmp = x[1][i] - x[0][i]; 44 ans = max(ans, tmp); 45 for(int j = 0; j < sz; ++j) { 46 int k = j; 47 while(k + 1 < sz && !cmp(p[k], p[k + 1])) ++k; 48 for(int r = j; r <= k; ++r) { 49 ll id = p[r].id; 50 if(!vis[id]) { 51 vis[id] = 1; 52 tmp += x[1][id] - x[0][id]; 53 p[r].id = 0; 54 } 55 } 56 ans = max(ans, tmp); 57 for(int r = j; r <= k; ++r) { 58 ll id = p[r].id; 59 if(vis[id]) { 60 vis[id] = 0; 61 tmp -= x[1][id] - x[0][id]; 62 } 63 } 64 j = k; 65 } 66 } 67 } 68 printf("%lld\n", ans); 69 return 0; 70 }
合法的一定可以看成是中间一堆单个的形式
1 #include <bits/stdc++.h> 2 using namespace std; 3 int a[111]; 4 5 int main() { 6 int n, sum = 0; 7 scanf("%d", &n); 8 for(int i = 1; i <= n; ++i) { 9 scanf("%d", a + i); 10 sum += a[i]; 11 } 12 if(sum & 1) puts("no quotation"); 13 else { 14 int ans = 1; 15 for(int i = 100; i > 1; --i) { 16 int l = 1, r = n, usedl = 0, usedr = 0, ok = 1; 17 for(int j = i; j >= 1; --j) { 18 if(l == r) { 19 if(usedl + 2 * j > a[l]) {ok = 0; break;} 20 usedl = usedr += 2 * j; 21 } 22 else { 23 if(usedl + j > a[l]) {ok = 0; break;} 24 usedl += j; 25 if(usedl == a[l]) { 26 l++; 27 if(l == r) usedl = usedr; 28 else usedl = 0; 29 } 30 if(usedr + j > a[r]) {ok = 0; break;} 31 usedr += j; 32 if(l == r) usedl += j; 33 if(r > l && usedr == a[r]) r--, usedr = 0; 34 } 35 } 36 if(ok) {ans = i; break;} 37 } 38 if(ans == 1 && sum != 2) puts("no quotation"); 39 else printf("%d\n", ans); 40 } 41 return 0; 42 }
决策单调性于是分治
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 vector<int> g[2][5555], d[2][5555]; 5 ll sum[5555], f[5555][5555]; 6 7 const int inf = 1e9; 8 typedef pair<int, int> pii; 9 int dist[5555], sumd[5555]; 10 void dijkstra(vector<int> g[], vector<int> d[], int s) { 11 priority_queue<pii> pq; 12 for(int i = 0; i < 5555; ++i) dist[i] = inf; 13 dist[s] = 0; 14 pq.push(pii(0, s)); 15 while (!pq.empty()) { 16 pii tmp = pq.top(); 17 pq.pop(); 18 int x = tmp.second; 19 if (dist[x] < -tmp.first) continue; 20 for (int i = 0; i < g[x].size(); ++i) { 21 int to = g[x][i], cost = d[x][i]; 22 if (dist[to] > dist[x] + cost) { 23 dist[to] = dist[x] + cost; 24 pq.push(pii(-dist[to], to)); 25 } 26 } 27 } 28 } 29 30 void solve(int i, int l, int r, int pl, int pr) { 31 int p = -1; 32 int m = (l + r) / 2; 33 for(int j = pl; j <= min(pr, m - 1); ++j) { 34 ll tmp = f[i - 1][j] + (m - j - 1) * (sum[m] - sum[j]); 35 if(tmp < f[i][m]) f[i][m] = tmp, p = j; 36 } 37 if(m > l) solve(i, l, m - 1, pl, p); 38 if(m < r) solve(i, m + 1, r, p, pr); 39 } 40 41 int main() { 42 int n, b, s, r; 43 scanf("%d %d %d %d", &n, &b, &s, &r); 44 for(int i = 1; i <= r; ++i) { 45 int u, v, w; 46 scanf("%d %d %d", &u, &v, &w); 47 g[0][u].push_back(v), d[0][u].push_back(w); 48 g[1][v].push_back(u), d[1][v].push_back(w); 49 } 50 int S = b + 1; 51 for(int i = 0; i < 2; ++i) { 52 dijkstra(g[i], d[i], S); 53 for(int j = 1; j <= b; ++j) sumd[j] += dist[j]; 54 } 55 sort(sumd + 1, sumd + 1 + b); 56 for(int i = 1; i <= b; ++i) sum[i] = sum[i - 1] + sumd[i]; 57 for(int i = 0; i < 5555; ++i) 58 for(int j = 0; j < 5555; ++j) 59 f[i][j] = 1e18; 60 f[0][0] = 0; 61 for(int i = 1; i <= s; ++i) 62 solve(i, i, b, i - 1, b - 1); 63 printf("%lld\n", f[s][b]); 64 return 0; 65 }
把不能吃，能吃但不必须吃，必须吃的边界搞出来，必须吃个数必须小于等于1，如果没有必须吃就从能吃里拿一个最左的，没有能吃的就gg了
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 const int maxn = 1e5 + 10; 5 ll a[maxn], b[maxn], num[maxn]; 6 7 typedef pair<ll, ll> pll; 8 set<pll> cant, can; 9 set<int> must; 10 11 int main() { 12 int m, k, c, sum = 0; 13 scanf("%d %d", &m, &k); 14 for(int i = 1; i <= m; ++i) scanf("%lld", a + i), sum += a[i]; 15 for(int i = 1; i <= k; ++i) { 16 scanf("%lld", b + i); 17 num[b[i]] += 1; 18 } 19 for(int i = 1; i <= m; ++i) { 20 // n * (a[i] / sum) - 1 < num[i] < n * (a[i] / sum) + 1 21 // can't eat 22 // (n + 1) * (a[i] / sum) - 1 < num[i] < (n + 1) * (a[i] / sum) + 1 23 // (num[i] - 1) * sum / a[i] < n + 1 <= num[i] * sum / a[i] 24 // can eat 25 // (n + 1) * (a[i] / sum) - 1 < num[i] + 1 < (n + 1) * (a[i] / sum) + 1 26 // num[i] * sum / a[i] < n + 1 < (num[i] + 1) * sum / a[i] 27 // must eat 28 // (num[i] + 1) * sum / a[i] <= n + 1 < (num[i] + 2) * sum / a[i] 29 if(k + 1 <= (num[i] * sum) / a[i]) cant.insert(pll((num[i] * sum) / a[i], i)); 30 else if(k + 1 < ((num[i] + 1) * sum + a[i] - 1) / a[i]) can.insert(pll(((num[i] + 1) * sum + a[i] - 1) / a[i], i)); 31 else must.insert(i); 32 } 33 for(c = k + 1; c <= k + sum; ++c) { 34 while(!cant.empty()) { 35 auto it = cant.begin(); 36 if(c > (*it).first) { 37 int x = (*it).second; 38 cant.erase(*it); 39 can.insert(pll(((num[x] + 1) * sum + a[x] - 1) / a[x], x)); 40 } 41 else break; 42 } 43 while(!can.empty()) { 44 auto it = can.begin(); 45 if(c >= (*it).first) { 46 int x = (*it).second; 47 can.erase(pll(((num[x] + 1) * sum + a[x] - 1) / a[x], x)); 48 must.insert(x); 49 } 50 else break; 51 } 52 if(must.size() > 1) break; 53 if(must.empty()) { 54 if(can.empty()) break; 55 auto it = can.begin(); 56 int x = (*it).second; 57 can.erase(pll(((num[x] + 1) * sum + a[x] - 1) / a[x], x)); 58 must.insert(x); 59 } 60 auto it = must.begin(); 61 int x = *it; 62 must.erase(*it); 63 num[x] += 1; 64 cant.insert(pll((num[x] * sum) / a[x], x)); 65 } 66 if(c == k + sum + 1) puts("forever"); 67 else printf("%d\n", c - k - 1); 68 return 0; 69 }
快乐画图每一天
1 #include <bits/stdc++.h> 2 using namespace std; 3 char ti[24 * 60][7][22]; 4 char g[111][7][22]; 5 6 char ept[7][5] = { 7 "....", 8 "....", 9 "....", 10 "....", 11 "....", 12 "....", 13 "...." 14 }; 15 16 char col[7][4] = { 17 "...", 18 "...", 19 ".X.", 20 "...", 21 ".X.", 22 "...", 23 "..." 24 }; 25 26 char non[7][22] = { 27 ".XX...XX.....XX...XX.", 28 "X..X.X..X...X..X.X..X", 29 "X..X.X..X.X.X..X.X..X", 30 ".XX...XX.....XX...XX.", 31 "X..X.X..X.X.X..X.X..X", 32 "X..X.X..X...X..X.X..X", 33 ".XX...XX.....XX...XX." 34 }; 35 36 char digit[10][7][5] = { 37 38 // 0 39 ".XX.", 40 "X..X", 41 "X..X", 42 "....", 43 "X..X", 44 "X..X", 45 ".XX.", 46 47 // 1 48 "....", 49 "...X", 50 "...X", 51 "....", 52 "...X", 53 "...X", 54 "....", 55 56 // 2 57 ".XX.", 58 "...X", 59 "...X", 60 ".XX.", 61 "X...", 62 "X...", 63 ".XX.", 64 65 // 3 66 ".XX.", 67 "...X", 68 "...X", 69 ".XX.", 70 "...X", 71 "...X", 72 ".XX.", 73 74 // 4 75 "....", 76 "X..X", 77 "X..X", 78 ".XX.", 79 "...X", 80 "...X", 81 "....", 82 83 // 5 84 ".XX.", 85 "X...", 86 "X...", 87 ".XX.", 88 "...X", 89 "...X", 90 ".XX.", 91 92 // 6 93 ".XX.", 94 "X...", 95 "X...", 96 ".XX.", 97 "X..X", 98 "X..X", 99 ".XX.", 100 101 // 7 102 ".XX.", 103 "...X", 104 "...X", 105 "....", 106 "...X", 107 "...X", 108 "....", 109 110 // 8 111 ".XX.", 112 "X..X", 113 "X..X", 114 ".XX.", 115 "X..X", 116 "X..X", 117 ".XX.", 118 119 // 9 120 ".XX.", 121 "X..X", 122 "X..X", 123 ".XX.", 124 "...X", 125 "...X", 126 ".XX." 127 }; 128 129 inline void draw(char g[][22], int c, int w, char fig[][5]) { 130 for(int i = 0; i < 7; ++i) { 131 for(int j = c; j < c + w; ++j) { 132 g[i][j] = fig[i][j - c]; 133 } 134 } 135 } 136 137 inline void drawcol(char g[][22]) { 138 for(int i = 0; i < 7; ++i) { 139 for(int j = 9; j < 12; ++j) { 140 g[i][j] = col[i][j - 9]; 141 } 142 } 143 } 144 145 bool on[7][22], off[7][22], work[7][22]; 146 int cnt_on[7][22], cnt_off[7][22]; 147 bool check(char g[][22], char fig[][22]) { 148 for(int i = 0; i < 7; ++i) { 149 for(int j = 0; j < 21; ++j) { 150 if(fig[i][j] == '.' && g[i][j] == 'X') on[i][j] = true; 151 if(fig[i][j] == 'X' && g[i][j] == '.') off[i][j] = true; 152 } 153 } 154 } 155 156 int main() { 157 for(int h = 0; h < 24; ++h) { 158 for(int m = 0; m < 60; ++m) { 159 int t = 60 * h + m; 160 if(h < 10) draw(ti[t], 0, 4, ept); 161 else draw(ti[t], 0, 4, digit[h / 10]); 162 drawcol(ti[t]); 163 draw(ti[t], 5, 4, digit[h % 10]); 164 draw(ti[t], 12, 4, digit[m / 10]); 165 draw(ti[t], 17, 4, digit[m % 10]); 166 } 167 } 168 int n, cnt = 0; 169 scanf("%d", &n); 170 for(int i = 0; i < n; ++i) 171 for(int j = 0; j < 7; ++j) 172 scanf("%s", g[i][j]); 173 for(int st = 0; st < 24 * 60; ++st) { 174 memset(on, 0, sizeof(on)); 175 memset(off, 0, sizeof(off)); 176 for(int i = 0; i < n; ++i) { 177 check(g[i], ti[(st + i) % (24 * 60)]); 178 } 179 int ok = 1; 180 for(int i = 0; i < n; ++i) { 181 for(int j = 0; j < 7; ++j) { 182 for(int k = 0; k < 21; ++k) { 183 if(on[j][k] && g[i][j][k] == '.') ok = 0; 184 if(off[j][k] && g[i][j][k] == 'X') ok = 0; 185 } 186 } 187 } 188 if(ok) { 189 cnt += 1; 190 for(int i = 0; i < 7; ++i) { 191 for(int j = 0; j < 21; ++j) { 192 if(on[i][j]) cnt_on[i][j] += 1; 193 if(off[i][j]) cnt_off[i][j] += 1; 194 } 195 } 196 } 197 } 198 for(int i = 1; i < n; ++i) { 199 for(int j = 0; j < 7; ++j) { 200 for(int k = 0; k < 21; ++k) { 201 if(g[i][j][k] != g[i - 1][j][k]) work[j][k] = 1; 202 } 203 } 204 } 205 if(!cnt) puts("impossible"); 206 else { 207 for(int i = 0; i < 7; ++i) { 208 for(int j = 0; j < 21; ++j) { 209 if(non[i][j] == '.') putchar('.'); 210 else if(cnt_on[i][j] == cnt) putchar('1'); 211 else if(cnt_off[i][j] == cnt) putchar('0'); 212 else if(work[i][j] && !cnt_on[i][j] && !cnt_off[i][j]) putchar('W'); 213 else putchar('?'); 214 } 215 puts(""); 216 } 217 } 218 return 0; 219 }
距离从大到小枚举叶子，如果一个子树里最短的距离都比这个大，至少有一个河会更长，就放弃这条河直接把海水引过来
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 const ll inf = 1e18; 5 const int maxn = 1e6 + 10; 6 vector<int> g[maxn]; 7 char s[maxn][22]; 8 int f[maxn]; 9 ll d[maxn]; 10 11 ll dis[maxn], mi[maxn]; 12 void dfs(int x) { 13 mi[x] = inf; 14 for(auto y: g[x]) { 15 dis[y] = dis[x] + d[y]; 16 dfs(y), mi[x] = min(mi[x], mi[y]); 17 } 18 if(mi[x] == inf) mi[x] = 0; 19 mi[x] += d[x]; 20 } 21 22 bool cmpnd(int i, int j) { 23 return mi[i] > mi[j]; 24 } 25 26 bool cmplf(int i, int j) { 27 return dis[i] > dis[j]; 28 } 29 30 int lf[maxn], nd[maxn], vis[maxn], used[maxn], ans[maxn]; 31 int main() { 32 int n, m; 33 scanf("%d %d", &n, &m); 34 for(int i = 1; i <= n; ++i) { 35 scanf("%s %d %lld", s[m + i], f + m + i, d + m + i); 36 g[f[m + i]].push_back(m + i); 37 lf[m + i] = nd[m + i] = m + i; 38 } 39 for(int i = 1; i <= m; ++i) { 40 scanf("%d %lld", f + i, d + i); 41 g[f[i]].push_back(i); 42 nd[i] = i; 43 } 44 dfs(0); 45 sort(nd + 1, nd + n + m + 1, cmpnd); 46 sort(lf + m + 1, lf + n + m + 1, cmplf); 47 int p = 0, tmp = 1; 48 for(int i = m + 1; i <= n + m; ++i) { 49 while(p < n + m && mi[nd[p + 1]] > dis[lf[i]]) { 50 int t = nd[p + 1]; 51 while(t && !vis[t]) { 52 vis[t] = 1; 53 if(!used[t]) tmp++; 54 if(vis[f[t]] && !used[f[t]]) tmp--; 55 used[f[t]] = 1; 56 t = f[t]; 57 } 58 ++p; 59 } 60 ans[lf[i]] = tmp; 61 } 62 for(int i = m + 1; i <= n + m; ++i) printf("%s %d\n", s[i], ans[i]); 63 return 0; 64 }
把每条路时间当变量就是线性规划
1 #include <bits/stdc++.h> 2 using namespace std; 3 int g[33][33]; 4 5 const double eps = 1e-8; 6 int cmp(const double &x) { 7 return x < -eps ? -1 : x > eps; 8 } 9 const int maxn = 444, maxm = 888, inf = 1e9; 10 double a[maxm][maxn], b[maxm]; 11 double ca[maxm][maxn], cb[maxm]; 12 int idn[maxn], idm[maxm]; 13 int n, m; 14 void pivot(int x,int y) { 15 swap(idm[x], idn[y]); 16 double k = a[x][y]; 17 for (int j = 1; j <= n; ++j)a[x][j] /= k; 18 b[x] /= k; 19 a[x][y] = 1 / k; 20 for (int i = 0; i <= m; ++i) 21 if (i != x) { 22 k = a[i][y]; 23 if (!cmp(k))continue; 24 b[i] -= k * b[x]; 25 a[i][y] = 0; 26 for (int j = 1; j <= n; ++j)a[i][j] -= a[x][j] * k; 27 } 28 } 29 void simplex() { 30 idn[0] = inf; 31 while (1) { 32 int y = 0; 33 for (int j = 1; j <= n; ++j) 34 if (cmp(a[0][j]) > 0 && idn[j] < idn[y])y = j; 35 if (!y)break; 36 int x = 0; 37 for (int i = 1; i <= m; ++i) 38 if (cmp(a[i][y]) > 0) { 39 if (!x)x = i; 40 else { 41 int t = cmp(b[i] / a[i][y] - b[x] / a[x][y]); 42 if (t < 0 || (t == 0 && idm[i] < idm[x]))x = i; 43 } 44 } 45 if (!x) { 46 puts("Unbounded"); 47 exit(0); 48 } 49 pivot(x, y); 50 } 51 } 52 void init() { 53 idm[0] = inf; 54 idn[0] = inf; 55 while (1) { 56 int x = 0; 57 for (int i = 1; i <= m; ++i) 58 if (cmp(b[i]) < 0 && idm[i] < idm[x])x = i; 59 if (!x)break; 60 int y = 0; 61 for (int j = 1; j <= n; ++j) 62 if (cmp(a[x][j]) < 0 && idn[j] < idn[y])y = j; 63 if (!y) { 64 puts("Infeasible"); 65 exit(0); 66 } 67 pivot(x, y); 68 } 69 } 70 71 int N; 72 vector<int> get_path(int s, int t) { 73 if(s == t) return vector<int>(1, s); 74 for(int i = 0; i < N; ++i) { 75 if(g[s][i] && g[i][t] && g[s][i] + g[i][t] == g[s][t]) { 76 vector<int> pre = get_path(s, i); 77 vector<int> suf = get_path(i, t); 78 for(int j = 1; j < suf.size(); ++j) pre.push_back(suf[j]); 79 return pre; 80 } 81 } 82 return vector<int>{s, t}; 83 } 84 85 int id[33][33]; 86 int main() { 87 int R, Q; 88 scanf("%d", &N); 89 for(int i = 0; i < N; ++i) { 90 for(int j = 0; j < N; ++j) { 91 scanf("%d", &g[i][j]); 92 if(i != j && g[i][j] != -1) { 93 id[i][j] = ++n; 94 ++m; 95 a[m][n] = 1, b[m] = g[i][j] + g[i][j]; 96 ++m; 97 a[m][n] = -1, b[m] = -g[i][j]; 98 } 99 if(g[i][j] == -1) g[i][j] = inf; 100 } 101 } 102 for(int k = 0; k < N; ++k) 103 for(int i = 0; i < N; ++i) 104 for(int j = 0; j < N; ++j) 105 g[i][j] = min(g[i][j], g[i][k] + g[k][j]); 106 scanf("%d", &R); 107 for(int i = 1; i <= R; ++i) { 108 int s, t, d; 109 scanf("%d %d %d", &s, &t, &d); 110 vector<int> path = get_path(s, t); 111 ++m; 112 for(int j = 0; j < path.size() - 1; ++j) { 113 a[m][id[path[j]][path[j + 1]]] = 1; 114 } 115 b[m] = d; 116 ++m; 117 for(int j = 0; j < path.size() - 1; ++j) { 118 a[m][id[path[j]][path[j + 1]]] = -1; 119 } 120 b[m] = -d; 121 } 122 memcpy(ca, a, sizeof(ca)); 123 memcpy(cb, b, sizeof(cb)); 124 scanf("%d", &Q); 125 while(Q--) { 126 int s, t; 127 scanf("%d %d", &s, &t); 128 printf("%d %d ", s, t); 129 vector<int> path = get_path(s, t); 130 memcpy(a, ca, sizeof(a)); 131 memcpy(b, cb, sizeof(b)); 132 for(int j = 1; j <= n; ++j) idn[j] = j; 133 for(int i = 1; i <= m; ++i) idm[i] = n + i; 134 for(int i = 0; i < path.size() - 1; ++i) a[0][id[path[i]][path[i + 1]]] = -1; 135 init(); 136 simplex(); 137 printf("%.9f ", b[0]); 138 memcpy(a, ca, sizeof(a)); 139 memcpy(b, cb, sizeof(b)); 140 for(int j = 1; j <= n; ++j) idn[j] = j; 141 for(int i = 1; i <= m; ++i) idm[i] = n + i; 142 for(int i = 0; i < path.size() - 1; ++i) a[0][id[path[i]][path[i + 1]]] = 1; 143 init(); 144 simplex(); 145 printf("%.9f\n", -b[0]); 146 } 147 return 0; 148 }
