题意：求点权最大的最短路，输出最短路径条数、点权值、以及最大点权的路径。
做法：Dijstra求最短路，两步：1.找最小点；2.更新路径。这题的1不变，变的是2。在更新新路径的时候，如果找到更短的路径，那么更新点权、方案数；如果路径和最短路径一样，那么比较谁的点权大，更新为点权大的结果。
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#include <bits/stdc++.h>#define scf0(a) scanf("%s",&a)#define scf1(a) scanf("%d",&a)#define scf2(a,b) scanf("%d%d",&a,&b)#define scf3(a,b,c) scanf("%d%d%d",&a,&b,&c)#define scf4(a,b,c,d) scanf("%d%d%d%d",&a,&b,&c,&d)#define MEM(a,b) memset(a,b,sizeof(a))#define pii pair<int,int>#define pdd pair<double,double>#define LL long longusing namespace std;const int maxn = 500 + 5;const int INF = 0x3f3f3f3f;int n, m, s, d;int u, v, len;int val[maxn];int mp[maxn][maxn];void Dijstra(){ bool vis[maxn]; MEM(vis, false); int dist[maxn]; //当前点的最短距离 MEM(dist, INF); int tot_val[maxn]; //当前点的最大点权 MEM(tot_val, 0); tot_val[s] = val[s]; dist[s] = 0; vis[s] = true; int diff[maxn]; //当前点不同最短路径方案数 for(int i = 0; i <= n; i++) diff[i] = 1; //方案数都为1 map<int,int>fa; //记录路径 fa.clear(); fa[s] = -1; for(int i = 0; i < n; i++) { if(i != s && dist[i] < INF) { dist[i] = mp[s][i]; tot_val[i] = val[s] + val[i]; fa[i] = s; } } for(int k = 0; k < n - 1; k++){ //找 n-1 个点 int minn = INF, pos; for(int i = 0; i < n; i++) { if(vis[i] == false) { if(dist[i] < minn) { minn = dist[i]; pos = i; } } } vis[pos] = true;// cout << pos << endl; for(int i = 0; i < n; i++) { if(vis[i] == false) { if(dist[pos] + mp[pos][i] < dist[i]) { dist[i] = dist[pos] + mp[pos][i]; tot_val[i] = tot_val[pos] + val[i]; fa[i] = pos; diff[i] = diff[pos]; //开始写成diff[i] = 1，WA了7分，应该继pos的路径数才对啊！ } else if(dist[pos] + mp[pos][i] == dist[i]) { diff[i] += diff[pos]; //开始写成diff[i]++， WA了7分，应该继pos的路径数才对啊！ if(tot_val[i] < tot_val[pos] + val[i]){ tot_val[i] = tot_val[pos] + val[i]; fa[i] = pos; } } } } } //输出 cout << diff[d] << ' '; stack<int>sta; int sum = val[d]; sta.push(d); while(fa[d] != -1) { d = fa[d]; sum += val[d]; sta.push(d); } cout << sum << endl; bool cnt = false; while(!sta.empty()) { if(cnt) cout << ' ';cnt = true; cout << sta.top(); sta.pop(); } cout << endl;}int main() { MEM(mp, INF); scf4(n, m, s, d); for(int i = 0; i < n; i++) scf1(val[i]); for(int i = 0; i < m; i++) { scf3(u, v, len); mp[u][v] = min(mp[u][v], len); //数据最终不卡相同路径值，但要注意这一点 mp[v][u] = mp[u][v]; } Dijstra();} |
无FUCK说...模模模模模拟....
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#include <bits/stdc++.h>#define scf0(a) scanf("%s",&a)#define scf1(a) scanf("%d",&a)#define scf2(a,b) scanf("%d%d",&a,&b)#define scf3(a,b,c) scanf("%d%d%d",&a,&b,&c)#define scf4(a,b,c,d) scanf("%d%d%d%d",&a,&b,&c,&d)#define MEM(a,b) memset(a,b,sizeof(a))#define PII pair<int,int>#define PDD pair<double,double>#define LL long longusing namespace std;struct Node { int key, next;}node[100100], node1[100100], node2[100100];int main() { int add0, n; scf2(add0,n); int x, y, z; for(int i = 0; i < n; i++) { scf3(x, y, z); node[x].key = y; node[x].next = z; } map<int,int>mp; int add1 = -1, add2 = -1; int first1 = -1, first2 = -1; bool cnt1 = false, cnt2 = false; while(add0 != -1) { if(!mp.count(abs(node[add0].key))) { if(!cnt1) first1 = add0; if(cnt1) node1[add1].next = add0; cnt1 = true; add1 = add0; node1[add1].key = node[add0].key; mp[abs(node[add0].key)] = 1; } else { if(!cnt2) first2 = add0; if(cnt2) node2[add2].next = add0; cnt2= true; add2 = add0; node2[add2].key = node[add0].key; } add0 = node[add0].next; } if(add1 != -1) node1[add1].next = -1; if(add2 != -1) node2[add2].next = -1; while(1) { if(first1 == -1) break; else if(node1[first1].next == -1) { printf("%05d %d -1
",first1, node1[first1].key); } else { printf("%05d %d %05d
",first1, node1[first1].key, node1[first1].next); } first1 = node1[first1].next; } while(1) { if(first2 == -1) break; else if(node2[first2].next == -1) { printf("%05d %d -1
",first2, node2[first2].key); } else { printf("%05d %d %05d
",first2, node2[first2].key, node2[first2].next); } first2 = node2[first2].next; }}//代码丑，有时间再改一下 |
无FUCK说
做法：STL的set搞定一切.....st.find(*it) A 了，find(st.begin(), st.end(), *it) T 了....
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#include <bits/stdc++.h>#define scf0(a) scanf("%s",&a)#define scf1(a) scanf("%d",&a)#define scf2(a,b) scanf("%d%d",&a,&b)#define scf3(a,b,c) scanf("%d%d%d",&a,&b,&c)#define scf4(a,b,c,d) scanf("%d%d%d%d",&a,&b,&c,&d)#define MEM(a,b) memset(a,b,sizeof(a))#define PII pair<int,int>#define PDD pair<double,double>#define LL long longusing namespace std;int main() { int n, x, y; scf1(n); set<int>st[100]; for(int i = 1; i <= n; i++) { scf1(x); while(x--) { scf1(y); st[i].insert(y); } } scf1(n); while(n--) { scf2(x, y); int c = 0, t = st[x].size() + st[y].size(); for(set<int>::iterator it = st[x].begin(); it != st[x].end(); it++) { if( st[y].find(*it) != st[y].end() ) //st.find(*it) 效率比algorithm里的find高 c++, t--; } double res = 100.0 * c / t; printf("%.2f%%
",res); }} |
题意：有N个人，每个人有不同的兴趣，只要有部分兴趣相同，那么就可以构成一个集合。
做法：这题挺简单的。就是并查集。
每个人的第一个兴趣的fa作为标准，将其他兴趣并进来，同时维护sum值（该集合的人数）。
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#include <bits/stdc++.h>#define scf1(a) scanf("%d", &a)#define scf2(a,b) scanf("%d%d", &a, &b)#define MEM(a,b) memset(a, b, sizeof(a))#define PII pair<int,int>using namespace std;const int INF = 0x3f3f3f3f;int fa[1100]; //dfs and similarint sum[1100]; //the sum of each groupint find(int x) { return fa[x] == x ? x : fa[x] = find(fa[x]);}int main() { int n, m, temp; scf1(n); for(int i = 1; i <= 1000; i++) fa[i] = i, sum[i] = 0; for(int j = 1; j <= n; j++) { scanf("%d: ", &m); scanf("%d", &temp); int x = find(temp); sum[x]++; for(int i = 2; i <= m; i++) { scf1(temp); int y = find(temp); if(x != y) { fa[y] = x; sum[x] += sum[y]; // y集合里的人数复制给x } } } multiset< int,greater<int> >ans; //开始用set WA了一个点 ans.clear(); for(int i = 1; i <= 1000; i++) { int x = find(i); if(x == i && sum[x] != 0) ans.insert(sum[x]); //因为没用到的点x也等于i，所以要判断sum[x]是否等于0 } cout << ans.size() << endl; set<int>::iterator it = ans.begin(); for(it; it != ans.end(); it++) { if(it != ans.begin()) cout << ' '; cout << *it; } cout << endl;} |
题意： 3D版的求联通块。每个联通块的个数>t的时候累加。
做法：BFS。还是挺简单的题，主要是读懂题。开始不大懂，看了样例数了一下感觉是这样做，结果1A...
（刚开始看不懂，把所有1加起来骗到了4分..2333）
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#include <bits/stdc++.h>#define scf1(a) scanf("%d", &a)#define scf2(a,b) scanf("%d%d", &a, &b)#define scf3(a,b,c) scanf("%d%d%d", &a, &b, &c)#define scf4(a,b,c,d) scanf("%d%d%d%d", &a, &b, &c, &d) #define MEM(a,b) memset(a, b, sizeof(a))#define PII pair<int,int>using namespace std;const int INF = 0x3f3f3f3f;int mp[100][1300][150];bool vis[100][1300][150];int fx[] = {0,0,0,0,1,-1};int fy[] = {0,0,-1,1,0,0};int fz[] = {-1,1,0,0,0,0};struct Node { int x, y, z;}node, temp;int main() { int idx = 0; int m, n, l, t; MEM(vis, false); scf4(m, n, l, t); for(int i = 0; i < l; i++) { for(int j = 0; j < m; j++) { for(int k = 0; k < n; k++) { scf1(mp[i][j][k]); } } } queue<Node>que; int ans = 0; for(int i = 0; i < l; i++) { for(int j = 0; j < m; j++) { for(int k = 0; k < n; k++) { if(vis[i][j][k] == true || mp[i][j][k] == 0) continue; vis[i][j][k] = true; while(!que.empty()) que.pop(); node.x = i, node.y = j, node.z = k; int sum = 0; //累计这个联通块包含几个点 que.push(node); while(!que.empty()) { sum++; node = que.front(); que.pop(); for(int t = 0; t < 6; t++) { temp.x = node.x + fx[t]; temp.y = node.y + fy[t]; temp.z = node.z + fz[t]; if(temp.x >= l || temp.x < 0 || temp.y >= m || temp.y < 0 || temp.z >= n || temp.z < 0) continue; if(vis[temp.x][temp.y][temp.z] == false && mp[temp.x][temp.y][temp.z] == 1) { vis[temp.x][temp.y][temp.z] = true; que.push(temp); } } } if(sum >= t) ans += sum; } } } cout << ans << endl;} |