以后每做完一场CF，解题报告都写在一起吧
暴力||二分 A - Bear and Elections
题意：有n个候选人，第一个候选人可以贿赂其他人拿到他们的票，问最少要贿赂多少张票第一个人才能赢
分析：正解竟然是暴力！没敢写暴力，卡了很久，导致这场比赛差点爆零！二分的话可以优化，但对于这题来说好像不需要。。。
收获：以后CF div2的A题果断暴力
代码(暴力)：
/************************************************
* Author :Running_Time
* Created Time :2015-8-30 0:40:46
* File Name :A.cpp
************************************************/
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e3 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int a[N];
int n, x;
int main(void) {
scanf ("%d", &n);
scanf ("%d", &a[0]); n--;
for (int i=1; i<=n; ++i) scanf ("%d", &a[i]);
int ans = 0;
while (true) {
int mxi = 0;
for (int i=1; i<=n; ++i) {
if (a[i] >= a[mxi]) mxi = i;
}
if (mxi == 0) break;
ans++; a[0]++; a[mxi]--;
}
printf ("%d\n", ans);
return 0;
}
代码(二分):
/************************************************
* Author :Running_Time
* Created Time :2015-8-30 0:40:46
* File Name :A.cpp
************************************************/
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e3 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int a[N];
int n;
bool cmp(int x, int y) {
return x > y;
}
bool check(int add) {
int y = a[0] + add;
for (int i=1; i<=n; ++i) {
if (a[i] >= y) {
if (add <= 0) return false;
add -= (a[i] - (y - 1));
if (add < 0) return false;
}
else break;
}
return true;
}
int main(void) {
scanf ("%d", &n);
scanf ("%d", &a[0]); n--;
for (int i=1; i<=n; ++i) scanf ("%d", &a[i]);
sort (a+1, a+1+n, cmp);
if (a[0] > a[1]) {
puts ("0"); return 0;
}
int l = 0, r = 1000;
while (l <= r) {
int mid = (l + r) >> 1;
if (check (mid)) r = mid - 1;
else l = mid + 1;
}
printf ("%d\n", l);
return 0;
}
暴力 B - Bear and Three Musketeers
题意：找一个三元环并且三个点的度数和-6最小
分析：三元环做过一题。这题能暴力跑过，DFS不用写。枚举边的两个端点，再找是否存在另外一个点构成三元环就可以了
收获：CF div2 B也暴力过
代码：
/************************************************
* Author :Running_Time
* Created Time :2015-8-25 19:24:24
* File Name :E_topo.cpp
************************************************/
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
typedef pair<int, int> PII;
const int N = 4e3 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int deg[N];
bool g[N][N];
vector<PII> G;
int n, m;
int main(void) {
scanf ("%d%d", &n, &m);
G.clear ();
memset (g, false, sizeof (g));
memset (deg, 0, sizeof (deg));
for (int u, v, i=1; i<=m; ++i) {
scanf ("%d%d", &u, &v);
G.push_back (PII (u, v));
g[u][v] = true; g[v][u] = true;
deg[u]++; deg[v]++;
}
int ans = INF;
for (int i=0; i<G.size (); ++i) {
int u = G[i].first, v = G[i].second;
for (int j=1; j<=n; ++j) {
if (g[u][j] && g[v][j]) {
ans = min (ans, deg[u] + deg[v] + deg[j] - 6);
}
}
}
printf ("%d\n", (ans == INF) ? -1 : ans);
return 0;
}
Codeforces Round #318 [RussianCodeCup Thanks-Round] (Div. 2)
原文：http://www.cnblogs.com/Running-Time/p/4771648.html