poj 3259

还是找带负权的环路

没找到自己写的问题在哪。。。

转的AC的http://blog.csdn.net/xsbailong/article/details/6864693

邻接矩阵对于松弛效率比较低，应该用边的结构体

//用邻接矩阵的一种实现，效率很低。。#include <iostream>#include <cstdio>#include <string.h>#include <limits.h>using namespace std;int map[510][510];int dist[510];void bellman( int n ){    int i,j,k;    bool in;    memset( dist,0,sizeof(dist) );    for( i=1;i<=n;i++ )    {        in=false;        for( j=1;j<=n;j++ )            for( k=1;k<=n;k++ )            {                if( map[j][k]!=0 )                {                    if( dist[j]!=INT_MAX &&  dist[k]>dist[j]+map[j][k] )                        dist[k]=dist[j]+map[j][k] , in=true ;                }            }        if( !in )   break;    }    if( i==n+1 )    printf( "YES\n" );    else printf( "NO\n" );}int main(){    int T,n,m,w,i,j;    int a,b,value;    scanf( "%d",&T );    while( T-- )    {        memset( map,0,sizeof(map) );        scanf( "%d%d%d",&n,&m,&w );        for( i=0;i<m;i++ )        {            scanf( "%d%d%d",&a,&b,&value );            if( map[a][b]==0 || map[a][b]>value )            {                map[a][b]=value;                map[b][a]=map[a][b];            }        }        for( i=0;i<w;i++ )        {            scanf( "%d%d%d",&a,&b,&value );            if( map[a][b]==0 || map[a][b]>-value )                map[a][b]=-value;        }        bellman(n);    }    return 0;}

我的怎么就不对呢。。。WAWAWA

#include <iostream>#include <sstream>#include <cstdio>#include <cstring>#include <cmath>#include <string>#include <vector>#include <set>#include <cctype>#include <algorithm>#include <cmath>#include <deque>#include <queue>#include <map>#include <queue>#include <list>#include <iomanip>using namespace std;                                               //////#define INF 20000#define maxn 2550#define max(a,b)(a>b?a:b)///int N, M, W;int maze[maxn][maxn];int dist[maxn];bool bellman(){int i, j, k;int lines;//总的边数lines = N;//初始化从0出发的最短距离数组distdist[1] = 0;for (i = 2; i <= N; i++){if (maze[1][i] != 0)dist[i] = maze[1][i];}//进行边数次松弛操作bool flag;for (i = 0; i < lines; i++){flag = false;for (j = 1; j <= N; j++){for (k = 1; k <= N; k++){if (maze[j][k] != 0){if (dist[j] != INF && dist[k] > dist[j] + maze[j][k]){dist[k] = dist[j] + maze[j][k];flag = true;}}}}if (!flag)break;}/*for (j = 1; j <= N; j++){for (k = 1; k <= N; k++){if (maze[j][k] != 0){if (dist[k] > dist[j] + maze[j][k])return true;}}}*/if (i == lines + 1)return true;elsereturn false;}int main(){freopen("F:\\input.txt","r",stdin );///int i, j;int cases;cin >> cases;while (cases--){memset(maze, 0, sizeof(maze));memset(dist, INF, sizeof(dist));cin >> N >> M >> W;//双向路径for (i = 0; i < M; i++){int  from, to ,weight;cin >> from >> to >> weight;maze[from][to] = weight;maze[to][from] = weight;}//单向负边路径for (i = 0; i < W; i++){int from, to, weight;cin >> from >> to >> weight;maze[from][to] = -weight;}if (bellman())cout << "YES" << endl;elsecout << "NO" << endl;}///    return 0;}