本文主要是介绍HDU/杭电2013多校第二场解题报告，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
这场一直在写1001，一直TLE。题解说的是DP，dp的用处.....
1001：就是变换球的位置求话费；想法太简单了，还是得用优化的算法，一直在写，一直T：
#include <cstdio>
#include <cstring>
#include <vector>
#include<queue>
#include<cmath>
#include<algorithm>
using namespace std;
#define MAXN 200006
#define MAXM 2000006
int n,a,b;
int main() {int ca;scanf("%d",&ca);while(ca--){scanf("%d%d%d",&n,&a,&b);int ans=0;if(a>b)swap(a,b);if(n<=a){puts("0");}else if(n>a&&n<b){for(int i=a;i<n;i++){ans+=abs(i%a-i%b);}}else{for(int i=a;i<n;i++){ans+=abs(i%a-i%b);}}printf("%d\n",ans);}return 0;
}
自己一直搞1001，后面的是其它的队友搞的
：
Palindrome Sub-Array
就是相当于求二维的回文，多条件判断：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <algorithm>
#include <vector>
#include <queue>
#include <map>
using namespace std;int s[305][305];
int n,m;
int l;
int ans;
void init()
{scanf("%d%d",&n,&m);for(int i=0; i<n; i++){for(int j=0; j<m; j++){scanf("%d",&s[i][j]);}}
}
void solve()
{ans = 1;for(int i=0; i<n; i++){for(int j=0; j<m; j++){if(min(n-i,m-j) <= ans){break;}l = min(n-i,m-j);while(l > ans){if(s[i][j]==s[i+l-1][j]&&s[i+l-1][j]==s[i][j+l-1]&&s[i][j+l-1]==s[i+l-1][j+l-1]){int q = 0;for(int k = 1; k * 2 < l; k++){if(s[i+k][j]==s[i+l-1-k][j]&&s[i+l-1-k][j]==s[i+k][j+l-1]&&s[i+k][j+l-1]==s[i+l-1-k][j+l-1]){if(s[i][j+k]==s[i+l-1][j+k]&&s[i+l-1][j+k]==s[i][j+l-1-k]&&s[i][j+l-1-k]==s[i+l-1][j+l-1-k]){if(s[i+k][j+k]==s[i+l-1-k][j+k]&&s[i+l-1-k][j+k]==s[i+k][j+l-1-k]&&s[i+k][j+l-1-k]==s[i+l-1-k][j+l-1-k])continue;else{q = 1;break;}}else{q = 1;break;}}else{q = 1;break;}for(int c = k - 1; c > 0; c--){if(s[i+k][j+c]==s[i+l-1-k][j+c]&&s[i+l-1-k][j+c]==s[i+k][j+l-1-c]&&s[i+k][j+l-1-c]==s[i+l-1-k][j+l-1-c]){if(s[i+c][j+k]==s[i+l-1-c][j+k]&&s[i+l-1-c][j+k]==s[i+c][j+l-1-k]&&s[i+c][j+l-1-k]==s[i+l-1-c][j+l-1-k]){continue;}else{q = 1;break;}}else{q = 1;break;}}if(q == 1)break;}if(q == 0)ans = l;}l--;}}}printf("%d\n",ans);
}
int main()
{int t;scanf("%d",&t);while(t--){init();solve();}return 0;
}
Warm up
#include <cstdio>
#include <cstring>
#include <vector>
#include<queue>
#include<cmath>
#include <algorithm>
using namespace std;
#define MAXN 200006
#define MAXM 2000006
#pragma comment(linker, "/STACK:1024000000,1024000000")
struct node {int v, w, pre;
} edge[MAXM];
int pos[MAXN], nEdge; //图的存储：链式前向星（池子法）struct Bridge {int u, v;
} bridge[MAXM]; //用来记录桥
int tot; //桥的个数int fa[MAXN], cc; //fa：各个点所属的缩点（连通块），cc连通块的个数
int dfn[MAXN], low[MAXN], Time; //时间戳
int stack[MAXN], top; //用于维护连通块的
int n, m; //点的个数和边的条数void connect(int u, int v, int w) {nEdge++;edge[nEdge].pre = pos[u];edge[nEdge].v = v;edge[nEdge].w = w;pos[u] = nEdge;
}
vector<int> graph[MAXN];
void tarjan(int cur, int from, int from_edge) {low[cur] = dfn[cur] = Time++;stack[++top] = cur;for (int p=pos[cur]; p; p=edge[p].pre) {int v = edge[p].v;if (v == from&&abs(p-from_edge)<=1) continue; //注意一下这里if (!dfn[v]) {tarjan(v, cur, p);if (low[v] < low[cur]) low[cur] = low[v];if (low[v] > dfn[cur]) {bridge[tot].u = cur;bridge[tot++].v = v;cc++;graph[cc].clear();do {fa[stack[top]] = cc;} while (stack[top--] != v);}} else if (low[cur] > dfn[v]) low[cur] = dfn[v];}
}
int bfs(int x,int flag)
{queue<int> q;int y;q.push(x);memset(dfn,0,sizeof(dfn));memset(low,0,sizeof(low));low[x]=0;dfn[x]=1;while(q.size()>0){y=q.front();// printf("%d\n",y);q.pop();for(int i=0;i<graph[y].size();i++){int z=graph[y][i];if(!dfn[z]){dfn[z]=1;low[z]=low[y]+1;q.push(z);}}}if(flag)return y;return low[y];
}
int main() {while(scanf("%d%d", &n, &m),n+m){memset(pos, 0, sizeof(pos));nEdge = 0;int u, v, w;for (int i=0; i<m; i++) {scanf("%d%d", &u, &v);connect(u, v, 1);connect(v, u, 1);}memset(dfn, 0, sizeof(dfn));memset(fa, -1, sizeof(fa));cc = tot = 0;for (int i=1; i<=n; i++)if (!dfn[i]) {top = Time = 1;tarjan(i, -1, -3);++cc;graph[cc].clear();for (int j=1; j<=n; j++)if (dfn[j] && fa[j]==-1) fa[j] = cc;}for(int i=0;i<tot;i++){graph[fa[bridge[i].u]].push_back(fa[bridge[i].v]);graph[fa[bridge[i].v]].push_back(fa[bridge[i].u]);}int y=bfs(1,1);int ans2=bfs(y,0);// printf("%d %d\n",cc,ans2);printf("%d\n",cc-1-(ans2));}return 0;
}
Warm up 2
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstring>
#include <queue>
#include <set>
#include <vector>
#include <stack>
#include <map>
#include <iomanip>
#define PI acos(-1.0)
#define Max 2505
#define inf 1<<28
#define LL(x) ( x << 1 )
#define RR(x) ( x << 1 | 1 )
#define REP(i,s,t) for( int i = ( s ) ; i <= ( t ) ; ++ i )
#define ll long long
#define mem(a,b) memset(a,b,sizeof(a))
#define mp(a,b) make_pair(a,b)
#define PII pair<int,int>
using namespace std;inline void RD(int &ret) {char c;do {c = getchar();} while(c < '0' || c > '9') ;ret = c - '0';while((c=getchar()) >= '0' && c <= '9')ret = ret * 10 + ( c - '0' );
}
int Map[1111][1111] ;
int link[1111] ;
bool vis[1111] ;
int n , m ,s, v;
int dfs(int now)
{for (int i = 1 ;i <= m ;i ++){if(Map[now][i])if(!vis[i]){vis[i] = 1 ;if(link[i] == -1 || dfs(link[i])){link[i] = now ;return 1 ;}}}return 0 ;
}int x1[1111], y1[1111] ;
int x2[11111],y2[1111] ;
bool check(int i , int j){if(x1[i] <= x2[j] && x2[j] <= x1[i] + 1){if(y2[j] <= y1[i] && y2[j] + 1 >= y1[i]){return 1 ;}}return 0 ;
}
int main() {while(scanf("%d%d",&n,&m) , (n + m)){mem(Map,0) ;mem(link , -1) ;for (int i = 1 ; i <= n ;i ++ ){scanf("%d%d",&x1[i] ,&y1[i]) ;}for (int i = 1 ; i <= m ;i ++ ){scanf("%d%d",&x2[i] , &y2[i]) ;}for (int i = 1 ; i <= n ; i ++ ){for (int j = 1 ; j <= m ;j ++ ){Map[i][j] = check(i , j ) ;}}
// for (int i = 1 ; i <= n ;i ++ ){
// for (int j = 1 ; j <= m ;j ++ ){
// cout << Map[i][j]<< " " ;
// }
// cout << endl;
// }int ans = 0 ;for (int i = 1 ; i <= n ;i ++){mem(vis, 0 ) ;ans += dfs(i) ;}cout << n + m - ans << endl;}return 0 ;
}
今天的题目赛后数据有加强的，还得继续努力啊。1001已经TLE到跪了，请大神指点...
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